EXC_BAD_ACCESS when executing an arm blx rx - c

Here is the c-source code line which crashes on an armv7:
ret = fnPtr (param1, param2);
In the debugger, fnPtr has an address of 0x04216c00. When I disassemble at the pc where it's pointing at the statement above, here is what I get:
0x18918e: movw r0, #0x73c
0x189192: movt r0, #0x1
0x189196: add r0, r2
0x189198: ldr r0, [r0]
0x18919a: str r0, [sp, #0x20]
0x18919c: ldr r0, [sp, #0x20]
0x18919e: ldr r1, [sp, #0x28]
0x1891a0: ldr r2, [sp, #0x2c]
0x1891a2: str r0, [sp, #0x14]
0x1891a4: mov r0, r1
0x1891a6: mov r1, r2
0x1891a8: ldr r2, [sp, #0x14]
0x1891aa: blx r2
Now, when I disassemble the memory at address $r2 (=0x4216c00), I get what is seemingly valid code that should be executed without any problem:
(lldb) disassemble -s 0x4216c00 -C 10
0x4216c00: push {r4, r5, r6, r7, lr}
0x4216c04: add r7, sp, #0xc
0x4216c08: push {r8, r10, r11}
0x4216c0c: vpush {d8, d9, d10, d11, d12, d13, d14, d15}
0x4216c10: sub r7, r7, #0x280
0x4216c14: mov r6, r0
0x4216c18: bx r1
0x4216c1c: add r7, r7, #0x280
Yet what really happens is this:
EXC_BAD_ACCESS (code=2, address=0x4216c00)
Can anyone explain what is wrong and why the address is considered illegal?
Full disclosure: I am no assembly expert. The code compiled and linked is all c-code. Compiler is clang.

Check the value of r2 before calling executing blx instruction. It might be odd, telling the cpu that address is in thumb mode however from the listing it looks like in arm mode.
Try forcing clang to only arm mode by -mno-thumb to test this.

The EXC_BAD_ACCESS exception has two bits of data in it, the first is the "kern_return_t" number describing the access failure, and the second is the address accessed. In your case the code is 2, which means (from /usr/include/mach/kern_return.h):
#define KERN_PROTECTION_FAILURE 2
/* Specified memory is valid, but does not permit the
* required forms of access.
*/
Not sure why this is happening, sounds like you are trying to execute code that doesn't have the execute permission set. What does:
(lldb) image lookup -va 0x4216c00
say?
BTW, the exception types are in /usr/include/mach/exception_types.h, and if the codes have machine specific meanings, those will be in, e.g. /usr/include/mach/i386/exception.h) For ARM info you may have to look in the header in the Xcode SDK.

Related

ARM Thumb GCC Disassembled C. Caller-saved registers not saved and loading and storing same register immediately

Context: STM32F469 Cortex-M4 (ARMv7-M Thumb-2), Win 10, GCC, STM32CubeIDE; Learning/Trying out inline assembly & reading disassembly, stack managements etc., writing to core registers, observing contents of registers, examining RAM around stack pointer to understand how things work.
I've noticed that at some point, when I call a function, in the beginning of a called function, which received an argument, the instructions generated for the C function do "store R3 at RAM address X" followed immediately "Read RAM address X and store in RAM". So it's writing and reading the same value back, R3 is not changed. If it only had wanted to save the value of R3 onto the stack, why load it back then?
C code, caller function (main), my code:
asm volatile(" LDR R0,=#0x00000000\n"
" LDR R1,=#0x11111111\n"
" LDR R2,=#0x22222222\n"
" LDR R3,=#0x33333333\n"
" LDR R4,=#0x44444444\n"
" LDR R5,=#0x55555555\n"
" LDR R6,=#0x66666666\n"
" MOV R7,R7\n" //Stack pointer value is here, used for stack data access
" LDR R8,=#0x88888888\n"
" LDR R9,=#0x99999999\n"
" LDR R10,=#0xAAAAAAAA\n"
" LDR R11,=#0xBBBBBBBB\n"
" LDR R12,=#0xCCCCCCCC\n"
);
testInt = addFifteen(testInt); //testInt=0x03; returns uint8_t, argument uint8_t
Function call generates instructions to load function argument into R3, then move it to R0, then branch with link to addFifteen. So by the time I enter addFifteen, R0 and R3 have value 0x03 (testInt). So far so good. Here is what function call looks like:
testInt = addFifteen(testInt);
08000272: ldrb r3, [r7, #11]
08000274: mov r0, r3
08000276: bl 0x80001f0 <addFifteen>
So I go into addFifteen, my C code for addFifteen:
uint8_t addFifteen(uint8_t input){
return (input + 15U);
}
and its disassembly:
addFifteen:
080001f0: push {r7}
080001f2: sub sp, #12
080001f4: add r7, sp, #0
080001f6: mov r3, r0
080001f8: strb r3, [r7, #7]
080001fa: ldrb r3, [r7, #7]
080001fc: adds r3, #15
080001fe: uxtb r3, r3
08000200: mov r0, r3
08000202: adds r7, #12
08000204: mov sp, r7
08000206: ldr.w r7, [sp], #4
0800020a: bx lr
My primary interest is in 1f8 and 1fa lines. It stored R3 on stack and then loads freshly written value back into the register that still holds the value anyway.
Questions are:
What is the purpose of this "store register A into RAM X, next read value from RAM X into register A"? Read instruction doesn't seem to serve any purpose. Make sure RAM write is complete?
Push{r7} instruction makes stack 4-byte aligned instead of 8-byte aligned. But immediately after that instruction we have SP decremented by 12 (bytes), so it becomes 8-byte aligned again. Therefore, this behavior is ok. Is this statement correct? What if an interrupt happens between these two instructions? Will alignment be fixed during ISR stacking for the duration of ISR?
From what I read about caller/callee saved registers (very hard to find any sort of well-organized information on that, if you have good material, please, share a link), at least R0-R3 must be placed on stack when I call a function. However, it's easy to notice in this case that NONE of the registers were pushed on stack, and I verified it by checking memory around stack pointer, it would have been easy to notice 0x11111111 and 0x22222222, but they aren't there, and nothing is pushing them there. The values in R0 and R3 that I had before I called the function are simply gone forever. Why weren't any registers pushed on stack before function call? I would expect to have R3 0x33333333 when addFifteen returns because that's how it was before function call, but that value is casually overwritten even before branch to addFifteen. Why didn't GCC generate instructions to push R0-R3 onto the stack and only after that branch with link to addFifteen?
If you need some compiler settings, please, let me know where to find them in Eclipse (STM32CubeIDE) and what exactly you need there, I will happily provide them and add them to the question here.
uint8_t addFifteen(uint8_t input){
return (input + 15U);
}
What you are looking at here is unoptimized and at least with gnu the input and local variables get a memory location on the stack.
00000000 <addFifteen>:
0: b480 push {r7}
2: b083 sub sp, #12
4: af00 add r7, sp, #0
6: 4603 mov r3, r0
8: 71fb strb r3, [r7, #7]
a: 79fb ldrb r3, [r7, #7]
c: 330f adds r3, #15
e: b2db uxtb r3, r3
10: 4618 mov r0, r3
12: 370c adds r7, #12
14: 46bd mov sp, r7
16: bc80 pop {r7}
18: 4770 bx lr
What you see with r3 is that the input variable, input, comes in r0. For some reason, code is not optimized, it goes into r3, then it is saved in its memory location on the stack.
Setup the stack
00000000 <addFifteen>:
0: b480 push {r7}
2: b083 sub sp, #12
4: af00 add r7, sp, #0
save input to the stack
6: 4603 mov r3, r0
8: 71fb strb r3, [r7, #7]
so now we can start implementing the code in the function which wants to do math on the input function, so do that math
a: 79fb ldrb r3, [r7, #7]
c: 330f adds r3, #15
Convert the result to an unsigned char.
e: b2db uxtb r3, r3
Now prepare the return value
10: 4618 mov r0, r3
and clean up and return
12: 370c adds r7, #12
14: 46bd mov sp, r7
16: bc80 pop {r7}
18: 4770 bx lr
Now if I tell it not to use a frame pointer (just a waste of a register).
00000000 <addFifteen>:
0: b082 sub sp, #8
2: 4603 mov r3, r0
4: f88d 3007 strb.w r3, [sp, #7]
8: f89d 3007 ldrb.w r3, [sp, #7]
c: 330f adds r3, #15
e: b2db uxtb r3, r3
10: 4618 mov r0, r3
12: b002 add sp, #8
14: 4770 bx lr
And you can still see each of the fundamental steps in implementing the function. Unoptimized.
Now if you optimize
00000000 <addFifteen>:
0: 300f adds r0, #15
2: b2c0 uxtb r0, r0
4: 4770 bx lr
It removes all the excess.
number two.
Yes I agree this looks wrong, but gnu certainly does not keep the stack on an alignment at all times, so this looks wrong. But I have not read the details on the arm calling convention. Nor have I read to see what gcc's interpretation is. Granted they may claim a spec, but at the end of the day the compiler authors choose the calling convention for their compiler, they are under no obligation to arm or intel or others to conform to any spec. Their choice, and like the C language itself, there are lots of places where it is implementation defined and gnu implements the C language one way and others another way. Perhaps this is the same. Same goes for this saving of the incoming variable to the stack. We will see that llvm/clang does not.
number three.
r0-r3 and another register or two may be called caller saved, but the better way to think of them is volatile. The callee is free to modify them without saving them. It is not so much a case of saving the r0 register, but instead r0 represents a variable and you are managing that variable in functionally implementing the high level code.
For example
unsigned int fun1 ( void );
unsigned int fun0 ( unsigned int x )
{
return(fun1()+x);
}
00000000 <fun0>:
0: b510 push {r4, lr}
2: 4604 mov r4, r0
4: f7ff fffe bl 0 <fun1>
8: 4420 add r0, r4
a: bd10 pop {r4, pc}
x comes in in r0, and we need to preserve that value until after fun1() is called. r0 can be destroyed/modified by fun1(). So in this case they save r4, not r0, and keep x in r4.
clang does this as well
00000000 <fun0>:
0: b5d0 push {r4, r6, r7, lr}
2: af02 add r7, sp, #8
4: 4604 mov r4, r0
6: f7ff fffe bl 0 <fun1>
a: 1900 adds r0, r0, r4
c: bdd0 pop {r4, r6, r7, pc}
Back to your function.
clang, unoptimized also keeps the input variable in memory (stack).
00000000 <addFifteen>:
0: b081 sub sp, #4
2: f88d 0003 strb.w r0, [sp, #3]
6: f89d 0003 ldrb.w r0, [sp, #3]
a: 300f adds r0, #15
c: b2c0 uxtb r0, r0
e: b001 add sp, #4
10: 4770 bx lr
and you can see the same steps, prep the stack, store the input variable. Take the input variable do the math. Prepare the return value. Clean up, return.
Clang/llvm optimized:
00000000 <addFifteen>:
0: 300f adds r0, #15
2: b2c0 uxtb r0, r0
4: 4770 bx lr
Happens to be the same as gnu. Not expected that any two different compilers generate the same code, nor any expectation that any two versions of the same compiler generate the same code.
unoptimized, the input and local variables (none in this case) get a home on the stack. So what you are seeing is the input variable being put in its home on the stack as part of the setup of the function. Then the function itself wants to operate on that variable so, unoptimized, it needs to fetch that value from memory to create an intermediate variable (that in this case did not get a home on the stack) and so on. You see this with volatile variables as well. They will get written to memory then read back then modified then written to memory and read back, etc...
yes I agree, but I have not read the specs. End of the day it is gcc's calling convention or interpretation of some spec they choose to use. They have been doing this (not being aligned 100% of the time) for a long time and it does not fail. For all called functions they are aligned when the functions are called. Interrupts in arm code generated by gcc is not aligned all the time. Been this way since they adopted that spec.
by definition r0-r3, etc are volatile. The callee can modify them at will. The callee only needs to save/preserve them if IT needs them. In both the unoptimized and optimized cases only r0 matters for your function it is the input variable and it is used for the return value. You saw in the function I created that the input variable was preserved for later, even when optimized. But, by definition, the caller assumes these registers are destroyed by called functions, and called functions can destroy the contents of these registers and no need to save them.
As far as inline assembly goes, which is a different assembly language than "real" assembly language. I think you have a ways to go before being ready for that, but maybe not. After decades of constant bare metal work I have found zero real use cases for inline assembly, the cases I see are laziness avoiding allowing real assembly into the make system or ways to avoid writing real assembly language. I see it as a ghee whiz feature that folks use like unions and bitfields.
Within gnu, for arm, you have at least four incompatible assembly languages for arm. The not unified syntax real assembly, the unified syntax real assembly. The assembly language that you see when you use gcc to assemble instead of as and then inline assembly for gcc. Despite claims of compatibility clang arm assembly language is not 100% compatible with gnu assembly language and llvm/clang does not have a separate assembler you feed it to the compiler. Arms various toolchains over the years have completely incompatible assembly language to gnu for arm. This is all expected and normal. Assembly language is specific to the tool not the target.
Before you can get into inline assembly language learn some of the real assembly language. And to be fair perhaps you do, and perhaps quite well, and this question is about the discover of how compilers generate code, and how strange it looks as you find out that it is not some one to one thing (all tools in all cases generate the same output from the same input).
For inline asm, while you can specify registers, depending on what you are doing, you generally want to let the compiler choose the register, most of the work for inline assembly is not the assembly but the language that specific compiler uses to interface it...which is compiler specific, move to another compiler and the expectation is a whole new language to learn. While moving between assemblers is also a whole new language at least the syntax of the instructions themselves tend to be the same and the language differences are in everything else, labels and directives and such. And if lucky and it is a toolchain not just an assembler, you can look at the output of the compiler to start to understand the language and compare it to any documentation you can find. Gnus documentation is pretty bad in this case, so a lot of reverse engineering is needed. At the same time you are more likely to be successful with gnu tools over any other, not because they are better, in many cases they are not, but because of the sheer user base and the common features across targets and over decades of history.
I would get really good at interfacing asm with C by creating mock C functions to see which registers are used, etc. And/or even better, implement it in C, compile it, then hand modify/improve/whatever the output of the compiler (you do not need to be a guru to beat the compiler, to be as consistent, perhaps, but fairly often you can easily see improvements that can be made on the output of gcc, and gcc has been getting worse over the last several versions it is not getting better, as you can see from time to time on this site). Get strong in the asm for this toolchain and target and how the compiler works, and then perhaps learn the gnu inline assembly language.
I'm not sure there is a specific purpose to do it. it is just one solution that the compiler has found to do it.
For example the code:
unsigned int f(unsigned int a)
{
return sqrt(a + 1);
}
compiles with ARM GCC 9 NONE with optimisation level -O0 to:
push {r7, lr}
sub sp, sp, #8
add r7, sp, #0
str r0, [r7, #4]
ldr r3, [r7, #4]
adds r3, r3, #1
mov r0, r3
bl __aeabi_ui2d
mov r2, r0
mov r3, r1
mov r0, r2
mov r1, r3
bl sqrt
...
and in level -O1 to:
push {r3, lr}
adds r0, r0, #1
bl __aeabi_ui2d
bl sqrt
...
As you can see the asm is much easier to understand in -O1: store parameter in R0, add 1, call functions.
The hardware supports non aligned stack during exception. See here
The "caller saved" registers do not necessarily need to be stored on the stack, it's up to the caller to know whether it needs to store them or not.
Here you are mixing (if I understood correctly) C and assembly: so you have to do the compiler job before switching back to C: either you store values in callee saved registers (and then you know by convention that the compiler will store them during function call) or you store them yourself on the stack.

Does arm-none-eabi-gcc produce slower code than Keil uVision

I have a simple blinking led program running on STM32f103C8 (without initialization boilerplate):
void soft_delay(void) {
for (volatile uint32_t i=0; i<2000000; ++i) { }
}
uint32_t iters = 0;
while (1)
{
LL_GPIO_TogglePin(LED_GPIO_Port, LED_Pin);
soft_delay();
++iters;
}
It was compiled with both Keil uVision v.5 (default compiler) and CLion using arm-none-eabi-gcc compiler.
The surprise is that arm-none-eabi-gcc program runs 50% slower in Release mode (-O2 -flto) and 100% slower in Debug mode.
I suspect 3 reasons:
Keil over-optimization (unlikely, because the code is very simple)
arm-none-eabi-gcc under-optimization due to wrong compiler flags (I use CLion Embedded plugins` CMakeLists.txt)
A bug in the initialization so that chip has lower clock frequency with arm-none-eabi-gcc (to be investigated)
I have not yet dived into the jungles of optimization and disassembling,
I hope that there are many experienced embedded developers who already encountered this issue and have the answer.
UPDATE 1
Playing around with different optimization levels of Keil ArmCC, I see
how it affects the generated code. And it affects drastically, especially execution time. Here are the benchmarks and disassembly of soft_delay() function for each optimization level (RAM and Flash amounts include initialization code).
-O0: RAM: 1032, Flash: 1444, Execution Time (20 iterations): 18.7 sec
soft_delay PROC
PUSH {r3,lr}
MOVS r0,#0
STR r0,[sp,#0]
B |L6.14|
|L6.8|
LDR r0,[sp,#0]
ADDS r0,r0,#1
STR r0,[sp,#0]
|L6.14|
LDR r1,|L6.24|
LDR r0,[sp,#0]
CMP r0,r1
BCC |L6.8|
POP {r3,pc}
ENDP
-O1: RAM: 1032, Flash: 1216, Execution Time (20 iterations): 13.3 sec
soft_delay PROC
PUSH {r3,lr}
MOVS r0,#0
STR r0,[sp,#0]
LDR r0,|L6.24|
B |L6.16|
|L6.10|
LDR r1,[sp,#0]
ADDS r1,r1,#1
STR r1,[sp,#0]
|L6.16|
LDR r1,[sp,#0]
CMP r1,r0
BCC |L6.10|
POP {r3,pc}
ENDP
-O2 -Otime: RAM: 1032, Flash: 1136, Execution Time (20 iterations): 9.8 sec
soft_delay PROC
SUB sp,sp,#4
MOVS r0,#0
STR r0,[sp,#0]
LDR r0,|L4.24|
|L4.8|
LDR r1,[sp,#0]
ADDS r1,r1,#1
STR r1,[sp,#0]
CMP r1,r0
BCC |L4.8|
ADD sp,sp,#4
BX lr
ENDP
-O3: RAM: 1032, Flash: 1176, Execution Time (20 iterations): 9.9 sec
soft_delay PROC
PUSH {r3,lr}
MOVS r0,#0
STR r0,[sp,#0]
LDR r0,|L5.20|
|L5.8|
LDR r1,[sp,#0]
ADDS r1,r1,#1
STR r1,[sp,#0]
CMP r1,r0
BCC |L5.8|
POP {r3,pc}
ENDP
TODO: benchmarking and disassembly for arm-none-eabi-gcc.
This second answer is a demonstration of the kinds of things that would affect the performance results the OP may be seeing and examples of to possibly test for those STM32F103C8 blue pill.
Complete source code:
flash.ld
MEMORY
{
rom : ORIGIN = 0x08000000, LENGTH = 0x1000
ram : ORIGIN = 0x20000000, LENGTH = 0x1000
}
SECTIONS
{
.text : { *(.text*) } > rom
.rodata : { *(.rodata*) } > rom
.bss : { *(.bss*) } > ram
}
flash.s
.cpu cortex-m0
.thumb
.thumb_func
.global _start
_start:
stacktop: .word 0x20001000
.word reset
.word hang
.word hang
.thumb_func
reset:
bl notmain
b hang
.thumb_func
hang: b .
.align
.thumb_func
.globl PUT32
PUT32:
str r1,[r0]
bx lr
.thumb_func
.globl GET32
GET32:
ldr r0,[r0]
bx lr
.thumb_func
.globl dummy
dummy:
bx lr
test.s
.cpu cortex-m0
.thumb
.word 0,0,0
.word 0,0,0,0
.thumb_func
.globl TEST
TEST:
bx lr
notmain.c
//PA9 TX
//PA10 RX
void PUT32 ( unsigned int, unsigned int );
unsigned int GET32 ( unsigned int );
void dummy ( unsigned int );
#define USART1_BASE 0x40013800
#define USART1_SR (USART1_BASE+0x00)
#define USART1_DR (USART1_BASE+0x04)
#define USART1_BRR (USART1_BASE+0x08)
#define USART1_CR1 (USART1_BASE+0x0C)
#define USART1_CR2 (USART1_BASE+0x10)
#define USART1_CR3 (USART1_BASE+0x14)
//#define USART1_GTPR (USART1_BASE+0x18)
#define GPIOA_BASE 0x40010800
#define GPIOA_CRH (GPIOA_BASE+0x04)
#define RCC_BASE 0x40021000
#define RCC_APB2ENR (RCC_BASE+0x18)
#define STK_CSR 0xE000E010
#define STK_RVR 0xE000E014
#define STK_CVR 0xE000E018
#define STK_MASK 0x00FFFFFF
static void uart_init ( void )
{
//assuming 8MHz clock, 115200 8N1
unsigned int ra;
ra=GET32(RCC_APB2ENR);
ra|=1<<2; //GPIOA
ra|=1<<14; //USART1
PUT32(RCC_APB2ENR,ra);
//pa9 TX alternate function output push-pull
//pa10 RX configure as input floating
ra=GET32(GPIOA_CRH);
ra&=~(0xFF0);
ra|=0x490;
PUT32(GPIOA_CRH,ra);
PUT32(USART1_CR1,0x2000);
PUT32(USART1_CR2,0x0000);
PUT32(USART1_CR3,0x0000);
//8000000/16 = 500000
//500000/115200 = 4.34
//4 and 5/16 = 4.3125
//4.3125 * 16 * 115200 = 7948800
PUT32(USART1_BRR,0x0045);
PUT32(USART1_CR1,0x200C);
}
static void uart_putc ( unsigned int c )
{
while(1)
{
if(GET32(USART1_SR)&0x80) break;
}
PUT32(USART1_DR,c);
}
static void hexstrings ( unsigned int d )
{
//unsigned int ra;
unsigned int rb;
unsigned int rc;
rb=32;
while(1)
{
rb-=4;
rc=(d>>rb)&0xF;
if(rc>9) rc+=0x37; else rc+=0x30;
uart_putc(rc);
if(rb==0) break;
}
uart_putc(0x20);
}
static void hexstring ( unsigned int d )
{
hexstrings(d);
uart_putc(0x0D);
uart_putc(0x0A);
}
void soft_delay(void) {
for (volatile unsigned int i=0; i<2000000; ++i) { }
}
int notmain ( void )
{
PUT32(STK_CSR,4);
PUT32(STK_RVR,0x00FFFFFF);
PUT32(STK_CVR,0x00000000);
PUT32(STK_CSR,5);
uart_init();
hexstring(0x12345678);
hexstring(GET32(0xE000E018));
hexstring(GET32(0xE000E018));
return(0);
}
build
arm-none-eabi-as --warn --fatal-warnings -mcpu=cortex-m3 flash.s -o flash.o
arm-none-eabi-as --warn --fatal-warnings -mcpu=cortex-m3 test.s -o test.o
arm-none-eabi-gcc -Wall -Werror -O2 -nostdlib -nostartfiles -ffreestanding -mthumb -mcpu=cortex-m0 -march=armv6-m -c notmain.c -o notmain.thumb.o
arm-none-eabi-ld -o notmain.thumb.elf -T flash.ld flash.o test.o notmain.thumb.o
arm-none-eabi-objdump -D notmain.thumb.elf > notmain.thumb.list
arm-none-eabi-objcopy notmain.thumb.elf notmain.thumb.bin -O binary
arm-none-eabi-gcc -Wall -Werror -O2 -nostdlib -nostartfiles -ffreestanding -mthumb -mcpu=cortex-m3 -march=armv7-m -c notmain.c -o notmain.thumb2.o
arm-none-eabi-ld -o notmain.thumb2.elf -T flash.ld flash.o test.o notmain.thumb2.o
arm-none-eabi-objdump -D notmain.thumb2.elf > notmain.thumb2.list
arm-none-eabi-objcopy notmain.thumb2.elf notmain.thumb2.bin -O binary
uart output as shown
12345678
00FFE445
00FFC698
If I take your code, make it shorter, don't have all day.
void soft_delay(void) {
for (volatile unsigned int i=0; i<0x2000; ++i) { }
}
arm-none-eabi-gcc -c -O0 -mthumb -mcpu=cortex-m0 hello.c -o hello.o
yes I know this is an m3
arm-none-eabi-gcc --version
arm-none-eabi-gcc (GCC) 5.4.0
gives
00000000 <soft_delay>:
0: b580 push {r7, lr}
2: b082 sub sp, #8
4: af00 add r7, sp, #0
6: 2300 movs r3, #0
8: 607b str r3, [r7, #4]
a: e002 b.n 12 <soft_delay+0x12>
c: 687b ldr r3, [r7, #4]
e: 3301 adds r3, #1
10: 607b str r3, [r7, #4]
12: 687b ldr r3, [r7, #4]
14: 4a03 ldr r2, [pc, #12] ; (24 <soft_delay+0x24>)
16: 4293 cmp r3, r2
18: d9f8 bls.n c <soft_delay+0xc>
1a: 46c0 nop ; (mov r8, r8)
1c: 46bd mov sp, r7
1e: b002 add sp, #8
20: bd80 pop {r7, pc}
22: 46c0 nop ; (mov r8, r8)
24: 00001fff
first check the test infrastructure
.cpu cortex-m0
.thumb
.align 8
.word 0,0
.thumb_func
.globl TEST
TEST:
push {r4,r5,r6,lr}
mov r4,r0
mov r5,r1
ldr r6,[r4]
inner:
bl soft_delay
sub r5,#1
bne inner
ldr r3,[r4]
sub r0,r6,r3
pop {r4,r5,r6,pc}
.align 8
soft_delay:
bx lr
in the openocd telnet window
reset halt
flash write_image erase notmain.thumb.elf
reset
gives
12345678
00001B59
7001 clocks, assuming the systick matches the cpu, thats 7001 arm clocks, 4 instructions per loop.
Step back note I aligned some things
08000108 <TEST>:
8000108: b570 push {r4, r5, r6, lr}
800010a: 1c04 adds r4, r0, #0
800010c: 1c0d adds r5, r1, #0
800010e: 6826 ldr r6, [r4, #0]
08000110 <inner>:
8000110: f000 f876 bl 8000200 <soft_delay>
8000114: 3d01 subs r5, #1
8000116: d1fb bne.n 8000110 <inner>
8000118: 6823 ldr r3, [r4, #0]
800011a: 1af0 subs r0, r6, r3
800011c: bd70 pop {r4, r5, r6, pc}
08000200 <soft_delay>:
8000200: 4770 bx lr
both loops are nicely aligned.
Now if I do this:
0800010a <TEST>:
800010a: b570 push {r4, r5, r6, lr}
800010c: 1c04 adds r4, r0, #0
800010e: 1c0d adds r5, r1, #0
8000110: 6826 ldr r6, [r4, #0]
08000112 <inner>:
8000112: f000 f875 bl 8000200 <soft_delay>
8000116: 3d01 subs r5, #1
8000118: d1fb bne.n 8000112 <inner>
800011a: 6823 ldr r3, [r4, #0]
800011c: 1af0 subs r0, r6, r3
800011e: bd70 pop {r4, r5, r6, pc}
Simply changing the alignment of the code that is supposed to be testing the code under test I now get:
00001F40
8000 ticks to do that loop 1000 times with that call with the code function under test still being aligned
08000200 <soft_delay>:
8000200: 4770 bx lr
The .align 8, in general don't use .align with a number on gnu its behavior does not translate across targets. .balign is better. Anyway I used it. The two words are because the align made TEST aligned, but inner is what I wanted aligned so I added two words to make it aligned.
.align 8
.word 0,0
nop
.thumb_func
.globl TEST
TEST:
push {r4,r5,r6,lr}
mov r4,r0
mov r5,r1
ldr r6,[r4]
inner:
bl soft_delay
sub r5,#1
bne inner
ldr r3,[r4]
sub r0,r6,r3
pop {r4,r5,r6,pc}
A little code review to make sure I didn't make a mistake here.
r0 is the systick current value register
r1 is the number of loops I want to run the code under test
The calling convention allows for r0-r3 to be clobbered so I need to move r0 and r1 to non-volatile registers (per the calling convention).
I want to sample the time the instruction before the loop and the instruction after.
so I need two registers for r0 and r1 and a register to store the begin time so r4,r5,r6 and that fits in nicely to have an even number of registers pushed on the stack. Have to preserve lr so we can return.
we can now safely call soft_delay in the loop, subtract the count, branch if not equal to inner, once the count is done read the timer in r3. from output above this is a down counter so subtract end from beginning, technically since this is a 24 bit counter I should and with 0x00FFFFFF to correctly do that subtraction, but because this isn't going to roll over I can assume out that operation. result/return value goes in r0, pop everything which includes popping the pc to do the return to the C calling function which prints out r0's value.
I think the test code is good.
reading the CPUID register
411FC231
So that means r1p1, while the TRM I am using is written for r2p1 you have to be very careful to use the right document but also sometimes use the current document or all the ones in between if available to see what changed.
ICode memory interface
Instruction fetches from Code memory space 0x00000000 to 0x1FFFFFFF
are performed over the 32-bit AHB-Lite bus. The Debugger cannot access
this interface. All fetches are word-wide. The number of instructions
fetched per word depends on the code running and the alignment of the
code in memory.
Sometimes in ARM TRMs you see the fetch info up top near the processor features, this tells me what I wanted to know.
08000112 <inner>:
8000112: f000 f875 bl 8000200 <soft_delay>
8000116: 3d01 subs r5, #1
8000118: d1fb bne.n 8000112 <inner>
this requires a fetch at 110, 114 and 118.
08000110 <inner>:
8000110: f000 f876 bl 8000200 <soft_delay>
8000114: 3d01 subs r5, #1
8000116: d1fb bne.n 8000110 <inner>
This a fetch at 110 and 114, but not one at 118, so that extra fetch could be our added clock. the m3 was the first publicly available one and it has a lot of features in the core that went away and similar ones came back. Some of the smaller cores fetch differently and you don't see this alignment issue. with bigger cores like full sized ones they fetch sometimes 4 or 8 instructions at a time and you have to change your alignment even more to hit the boundary but you can hit the boundary and since it is 2 or 4 clocks plus bus overhead for the extra fetch you can see those.
If I put two nops
nop
nop
.thumb_func
.globl TEST
TEST:
gives
08000114 <inner>:
8000114: f000 f874 bl 8000200 <soft_delay>
8000118: 3d01 subs r5, #1
800011a: d1fb bne.n 8000114 <inner>
800011c: 6823 ldr r3, [r4, #0]
800011e: 1af0 subs r0, r6, r3
8000120: bd70 pop {r4, r5, r6, pc}
gives
00001B59
So that's good we are back to that number, could try a few more to confirm but it appears that alignment is sensitive to our outer test loop, which is bad, but we can manage that, don't change it it won't affect the test. If I didn't care about alignment and had something like this:
void soft_delay(void) {
for (volatile unsigned int i=0; i<0x2000; ++i) { }
}
int notmain ( void )
{
unsigned int ra;
unsigned int beg;
unsigned int end;
PUT32(STK_CSR,4);
PUT32(STK_RVR,0x00FFFFFF);
PUT32(STK_CVR,0x00000000);
PUT32(STK_CSR,5);
uart_init();
hexstring(0x12345678);
beg=GET32(STK_CVR);
for(ra=0;ra<1000;ra++)
{
soft_delay();
}
end=GET32(STK_CVR);
hexstring((beg-end)&0x00FFFFFF);
return(0);
}
Then as I played with optimization options and I also played with using different compilers any change in the program/binary in front of the test loop would/could move the test loop changing its performance, in my simple example it was a 14% performance difference, that's massive if you are doing performance tests. letting the compiler take care of all this without us being in control the everything in front of the function under test could mess with the function under test, as written above the compiler might opt to inline the function rather than call it making an even more interesting situation as the test loop while probably not as clean as mine, certainly not if not optimized, but now the code under test is dynamic as options or alignments change.
I'm very happy you happened to be using this core/chip...
If I re-align inner and now mess with this
.align 8
nop
soft_delay:
bx lr
08000202 <soft_delay>:
8000202: 4770 bx lr
it's a single instruction which is fetched at 0x200 from what we have read and seem to be able to tell. wouldn't expect this to change anything and it didn't
00001B59
but now that we know what we know, we can use our experience to mess with this trivial Not interesting at all example.
.align 8
nop
soft_delay:
nop
bx lr
gives
00001F41
as expected. and we can have even more fun:
.align 8
.word 0,0
nop
.thumb_func
.globl TEST
TEST:
combined gives
08000112 <inner>:
8000112: f000 f876 bl 8000202 <soft_delay>
8000116: 3d01 subs r5, #1
8000118: d1fb bne.n 8000112 <inner>
08000202 <soft_delay>:
8000202: 46c0 nop ; (mov r8, r8)
8000204: 4770 bx lr
no surprise if you know what you are doing:
00002328
9000 clocks, 29% performance difference. we are literally talking about 5 (technically 6) instructions, same exact machine code and by simply changing alignment the performance can be 29% different, compiler and options have nothing to do with it, yet, have not even gotten there.
How can we expect to do any kind of performance evaluation of a program using the time the code a bunch of times in a loop method? We cant unless we know what we are doing, have an understanding of the architecture, etc.
Now as it should be obvious and reading the documentation I am using the internal 8Mhz clock, everything is derived from that so the systick times are not going to sometimes vary as you might see with dram for example. The LATENCY bits in the FLASH_ACR register should have defaulted to zero wait states for 0 < SYSCLK <- 24Mhz. If I were to bump up the clock above 24Mhz, the processor is running faster but the flash is now slower relative to the processor.
Without messing with the clocks and simply adding a wait state by changing the FLASH_ACR register to 0x31.
000032C6
12998 up from 9000, I didn't expect it to double necessarily and it didn't.
Hmm for fun make a PUT16 using strh, and
.thumb_func
.globl HOP
HOP:
bx r2
and
PUT16(0x2000010a,0xb570); // 800010a: b570 push {r4, r5, r6, lr}
PUT16(0x2000010c,0x1c04); // 800010c: 1c04 adds r4, r0, #0
PUT16(0x2000010e,0x1c0d); // 800010e: 1c0d adds r5, r1, #0
PUT16(0x20000110,0x6826); // 8000110: 6826 ldr r6, [r4, #0]
PUT16(0x20000112,0xf000); // 8000112: f000 f876 bl 8000202 <soft_delay>
PUT16(0x20000114,0xf876); // 8000112: f000 f876 bl 8000202 <soft_delay>
PUT16(0x20000116,0x3d01); // 8000116: 3d01 subs r5, #1
PUT16(0x20000118,0xd1fb); // 8000118: d1fb bne.n 8000112 <inner>
PUT16(0x2000011a,0x6823); // 800011a: 6823 ldr r3, [r4, #0]
PUT16(0x2000011c,0x1af0); // 800011c: 1af0 subs r0, r6, r3
PUT16(0x2000011e,0xbd70); // 800011e: bd70 pop {r4, r5, r6, pc}
PUT16(0x20000202,0x46c0); // 8000202: 46c0 nop ; (mov r8, r8)
PUT16(0x20000204,0x4770); // 8000204: 4770 bx lr
hexstring(HOP(STK_CVR,1000,0x2000010B));
gives
0000464B
and that was not at all expected. but is 18,000 basically
Putting ram to bed after this
PUT16(0x20000108,0xb570); // 800010a: b570 push {r4, r5, r6, lr}
PUT16(0x2000010a,0x1c04); // 800010c: 1c04 adds r4, r0, #0
PUT16(0x2000010c,0x1c0d); // 800010e: 1c0d adds r5, r1, #0
PUT16(0x2000010e,0x6826); // 8000110: 6826 ldr r6, [r4, #0]
PUT16(0x20000110,0xf000); // 8000112: f000 f876 bl 8000202 <soft_delay>
PUT16(0x20000112,0xf876); // 8000112: f000 f876 bl 8000202 <soft_delay>
PUT16(0x20000114,0x3d01); // 8000116: 3d01 subs r5, #1
PUT16(0x20000116,0xd1fb); // 8000118: d1fb bne.n 8000112 <inner>
PUT16(0x20000118,0x6823); // 800011a: 6823 ldr r3, [r4, #0]
PUT16(0x2000011a,0x1af0); // 800011c: 1af0 subs r0, r6, r3
PUT16(0x2000011c,0xbd70); // 800011e: bd70 pop {r4, r5, r6, pc}
PUT16(0x20000200,0x46c0); // 8000202: 46c0 nop ; (mov r8, r8)
PUT16(0x20000200,0x4770); // 8000204: 4770 bx lr
hexstring(HOP(STK_CVR,1000,0x20000109));
00002EDE
The machine code did not change because I moved both back by 2 so the relative address between them was the same. Note that bl is two separate instructions not one 32 bit one. You cant see this in the newer docs you need to go back to the original/early ARM ARM where it is explained. And it is easy to do experiments where you split the two instructions and put other stuff in between and they work just fine, because they are two separate instructions.
At this point the reader should be able to make a 2 instruction test loop, time it and dramatically change the performance of the execution of those two instructions on this platform using the same exact machine code.
So let's try the volatile loop that you wrote.
.align 8
soft_delay:
push {r7, lr}
sub sp, #8
add r7, sp, #0
mov r3, #0
str r3, [r7, #4]
b L12
Lc:
ldr r3, [r7, #4]
add r3, #1
str r3, [r7, #4]
L12:
ldr r3, [r7, #4]
ldr r2, L24
cmp r3, r2
bls Lc
nop
mov sp, r7
add sp, #8
pop {r7, pc}
nop
.align
L24: .word 0x1FFF
this is I believe the unoptimized -O0 version. starting off with one test loop
hexstring(TEST(STK_CVR,1));
experience, the times we are seeing will overflow our 24 bit counter and the results will be very strange or lead to false conclusions.
0001801F
98,000, quick check for safety:
.align
L24: .word 0x1F
0000019F
not bad that is on par with 256 times faster.
so we have some wiggle room in our test loop but not much try 10
hexstring(TEST(STK_CVR,10));
000F012D
98334 ticks per loop.
changing the alignment
08000202 <soft_delay>:
8000202: b580 push {r7, lr}
8000204: b082 sub sp, #8
gave the same result
000F012D
not unheard of, you can examine the differences if you want count through each instruction check fetch cycles, etc.
had I made the test:
soft_delay:
nop
nop
bx lr
its two fetch cycles no matter what the alignment or if I had left it bx lr with no nops as we saw so by simply having an odd number of instructions in the test then alignment won't affect the results on fetches along, but note that from what we know now had some other code in the program moved the outer timing/test loop that may have changed performance and the results may show a difference between two tests that were purely the timing code and not the code under test (read Michael Abrash).
The cortex-m3 is based on the armv7-m architecture. If I change the compiler from -mcpu=cortex-m0 (all cortex-m compatible so far) to -mcpu=cortex-m3 (not all cortex-m compatible will break on half of them) it produces a little bit less code.
.align 8
soft_delay:
push {r7}
sub sp, #12
add r7, sp, #0
movs r3, #0
str r3, [r7, #4]
b L12
Lc:
ldr r3, [r7, #4]
add r3, #1
str r3, [r7, #4]
L12:
ldr r3, [r7, #4]
/*14: f5b3 5f00 cmp.w r3, #8192 ; 0x2000*/
//cmp.w r3, #8192
.word 0x5f00f5b3
bcc Lc
nop
add r7, #12
mov sp, r7
pop {r7}
bx lr
000C80FB 81945 ticks for the code under test.
I hate unified syntax, that was a massive mistake, so I fumble along in legacy mode. thus the .word thing there in the middle.
As part of writing this I kinda messed up my system in order to demonstrate something. I was building a gcc 5.4.0 but overwrote my 9.2.0 so had to re-build both.
2.95 was the version I started using with arm and didn't support thumb gcc 3.x.x was the first to. And either gcc 4.x.x or gcc 5.x.x produced "slower" code for some of my projects, at work we are currently moving from ubuntu 16.04 to 18.04 for our build systems which if you use the apt-got cross compiler for arm that moves you from 5.x.x to 7.x.x and it is making larger binaries for the same source code and where we are tight on memory it is pushing us beyond what's available so we have to either remove some code (easiest to make the printed messages shorter, cut text out) or stick to the older compiler by building our own or apt-getting the older one. 19.10 does no longer offers the 5.x.x version.
So both are now built.
18: d3f8 bcc.n c <soft_delay+0xc>
1a: bf00 nop
1c: bf00 nop
1e: 370c adds r7, #12
these nops after bcc are baffling to me...
18: d3f8 bcc.n c <soft_delay+0xc>
1a: bf00 nop
1c: 370c adds r7, #12
gcc 5.4.0 is putting one, gcc 9.2.0 is putting two nops, ARM doesn't have the branch shadow thing of MIPS (MIPS doesn't currently either).
000C80FB gcc 5.4.0
000C8105 gcc 9.2.0
I call the function 10 times, the nop is outside the code under tests loop so has a lesser effect.
Optimized all cortex-m variants (to date) using gcc 9.2.0
soft_delay:
mov r3, #0
mov r2, #128
sub sp, #8
str r3, [sp, #4]
ldr r3, [sp, #4]
lsl r2, r2, #6
cmp r3, r2
bcs L1c
L10:
ldr r3, [sp, #4]
add r3, #1
str r3, [sp, #4]
ldr r3, [sp, #4]
cmp r3, r2
bcc L10
L1c:
add sp, #8
bx lr
(also understand that not all say gcc 9.2.0 builds produce the same code when you build the compiler you have options and those options can affect the output making different builds of 9.2.0 possibly producing different results)
000C80B5
gcc 9.2.0 built for cortex-m3:
soft_delay:
mov r3, #0
sub sp, #8
str r3, [sp, #4]
ldr r3, [sp, #4]
/*8: f5b3 5f00 cmp.w r3, #8192 ; 0x2000*/
.word 0x5F00F5B3
bcs L1c
Le:
ldr r3, [sp, #4]
add r3, #1
str r3, [sp, #4]
ldr r3, [sp, #4]
/*16: f5b3 5f00 cmp.w r3, #8192 ; 0x2000*/
.word 0x5F00F5B3
bcc Le
L1c:
add sp, #8
bx lr
000C80A1
That's in the noise. despite the code built has differences. they simply didn't gain in comparing the 0x2000 in fewer instructions. and note if you change that 0x2000 to some other number then that does not simply make the loop take that much longer it can change the generated code for architectures like this.
How I like to make these counted delay loops is to use a function outside the compile domain
extern void dummy ( unsigned int );
void soft_delay(void) {
for (unsigned int i=0; i<0x2000; ++i) { dummy(i); }
}
soft_delay:
push {r4, r5, r6, lr}
mov r5, #128
mov r4, #0
lsl r5, r5, #6
L8:
mov r0, r4
add r4, #1
bl dummy
cmp r4, r5
bne L8
pop {r4, r5, r6, pc}
the feature there is you don't need the overhead of what volatile does you do have a call and clearly there is overhead as well due to the call but not as much
000B40C9
or even better:
soft_delay:
sub r0,#1
bne soft_delay
bx lr
I would have to change the code wrapped around the code under test to make that function work.
Another note specific to these targets but also something you deal with
unsigned int more_fun ( unsigned int, unsigned int );
unsigned int fun ( unsigned int a, unsigned int b )
{
return(more_fun(a,b)+a+(b<<2));
}
00000000 <fun>:
0: b570 push {r4, r5, r6, lr}
2: 000c movs r4, r1
4: 0005 movs r5, r0
6: f7ff fffe bl 0 <more_fun>
a: 00a4 lsls r4, r4, #2
c: 1964 adds r4, r4, r5
e: 1820 adds r0, r4, r0
10: bd70 pop {r4, r5, r6, pc}
12: 46c0 nop ; (mov r8, r8)
a question repeated here at SO on a period basis. why is it pushing r6 it isn't using r6.
The compiler operates using what I call and used to be called a calling convention, now they use terms ABI, EABI, whatever either case it is the same thing it is a set of rules the compiler follows for a particular target. Arm added a rule to keep the stack aligned on a 64 bit address boundary instead of 32, this caused the extra item to keep the stack aligned, what register is used there can vary. If you use an older gcc vs a newer this can/will affect the performance of your code all by itself.
There are many factors at play here. Certainly if you have an optimizing compiler and you compare optimized vs not DEPENDING ON THE CODE you can see a large difference in execution speed. Using the volatile in the tiny loop here actually masks some of that, in both cases it should be read/written to memory every loop.
But the calling code the loop variable unoptimized would touch ram two or three times in that loop, optimized ideally would be in a register the whole time, making for a dramatic difference in execution performance even with zero wait state ram.
The toggle pin code is relatively large (talking to the peripheral directly would be less code), depending on whether that library was compiled separately with different options or at the same time with the same options makes a big difference with respect to performance.
Add that this is an mcu and running off of a flash which with the age of this part the flash might at best be half the clock rate of the cpu and worst a number of wait states and I don't remember off hand if ST had the caching in front of it at that time. so every instruction you add can add a clock, so just the loop variable alone can dramatically change the timing.
Being a high performance pipelined core I have demonstrated here and elsewhere that alignment can (not always) play a role, so if in one case the exact same machine code links to address 0x100 in one case and 0x102 in another it is possible that exact same machine code takes extra or fewer clocks to execute based on the nature of the pre-fetcher in the design, or the flash implementation, cache if any implementation, branch predictor, etc.
And then the biggest problem is how did you time this, it is not uncommon for there to be error in not using a clock correctly such that the clock/timing code itself varies and is creating some of the difference. Plus are there background things going on, interrupts/multitasking.
Michal Abrash wrote a wonderful book called The Zen of Assembly Language, you can get it for free in ePub or perhaps pdf form on GitHub. the 8088 was obsolete when the book was released but if you focus on that then you have missed the point, I bought it when it came out and have used what I learned on nearly a daily basis.
gcc is not a high performance compiler it is more of a general purpose compiler built Unix style where you can have different language front ends and different target backends. When I was in the position you are now trying to first understand these things I sampled many compilers for the same arm target and same C code and the results were vast. Later I wrote an instruction set simulator so I could count instructions and memory accesses to compare gnu vs llvm, as the latter has more optimization opportunities than gnu but for execution tests of code gcc was faster sometimes but not slower. That ended up being more of a long weekend having fun than something I used to analyze the differences.
It is easier to start with small-ish code like this and disassemble the two. Understand that fewer instructions doesn't mean faster, one distant memory access on a dram based system can take hundreds of clock cycles, that might be replaced with another solution that takes a handful/dozen of linearly fetched instructions to end up with the same result (do some math vs look up something in a rarely sampled table) and depending on the situation the dozen instructions execute much faster. at the same time the table solution can be much faster. it depends.
Examination of the disassembly often leads to incorrect conclusions (read abrash, not just that book, everything) as first off folks think less instructions means faster code. rearranging instructions in a pipelined processor can improve performance if you move an instruction into a time period that would have otherwise been wasted clocks. incrementing a register not related to a memory access in front of the memory access instead of after in a non-superscaler processor.
Ahh, back to a comment. This was years ago and competing compilers were more of a thing most folks just wrap their gui around gnu and the ide/gui is the product not the compiler. But there was the arm compiler itself, before the rvct tools, ads and I forget the other, those were "better" than gcc. I forget the names of the others but there was one that produced significantly faster code, granted this was Dhrystone so you will also find that they may tune optimizers for Dhrystone just to play benchmark games. Now that I can see how easy it is to manipulate benchmarks I consider them to in general be bu33zzit, can't be trusted. Kiel used to be a multi-target tool for mcus and similar, but then was purchased by arm and I thought at the time they were dropping all other targets, but have not checked in a while. I might have tried them once to get access to a free/demo version of rvct as when I was working at one job we had a budget to buy multi-thousand dollar tools, but that didn't include rvct (although I was on phone calls with the formerly Allant folks who were part of a purchase that became the rvct tools) which I was eager to try once they had finished the development/integration of that product, by then didn't have a budget for that and later couldn't afford or wasn't interested in buying even kiels tools, much less arms. Their early demos of rvct created an encrypted/obfuscated binary that was not arm machine code it only ran on their simulator so you couldn't use it to eval performance or to compare it to others, don't think they were willing to give us an un-obfuscated version and we weren't willing to reverse engineer it. now it is easier just to use gcc or clang and hand optimize where NEEDED. Likewise with experience can write C code that optimizes better based on experience examining compiler output.
You have to know the hardware, particularly in this case where you take processor IP and most of the chip is not related to the processor IP and most of the performance is not related to the processor IP (pretty much true for a lot of platforms today in particular your server/desktop/laptop). The Gameboy Advance for example used a lot of 16 bit buses instead of 32, thumb tools were barely being integrated, but thumb mode while counting instructions/or bytes was like 10% more code at the time, executed significantly faster on that chip. On other implementations both arm architecture and chip design thumb may have performance penalties, or not.
ST in general with the cortex-m products tends to put a cache in front of the flash, sometimes they document it and provide enable/disable control sometimes not so it can be difficult at best to get a real performance value as the typical thing is to run the code under test many times in a loop so you can get a better time measurement. other vendors don't necessarily do this and it is much easier to see the flash wait states and get a real, worst case, timing value that you can use to validate your design. caches in general as well as pipelines make it difficult at best to get good, repeatable, reliable numbers to validate your design. So for example you sometimes cannot do the alignment trick to mess with performance of the same machine code on an st but on say a ti with the same core you can. st might not give you the icache in a cortex-m7 where another vendor might since st has already covered that. Even within one brand name though don't expect the results of one chip/family to translate to another chip/family even if they use the same core. Also look at subtle comments in the arm documentation as to whether some cores offer a fetch size as an advertised option, single or multi-cycle multiply, etc. and I'll tell you that there are other compile time options for the core that are not shown in the technical reference manual that can affect performance so don't assume that all cortex-m3s are the same even if they are the same revision from arm. The chip vendor has the source so they can go even further and modify it, or for example a register bank to be implemented by the consumer they might change it from no protection to parity to ecc which might affect performance while retaining all of arms original code as is. When you look at an avr or pic though or even an msp430, while I cant prove it those designs appear more static not tiny vs Xmega vs regular old avr as there are definite differences there, but one tiny to another.
Your assumptions are a good start, there really isn't such a thing as over-optimization, more of a thing of missed optimizations, but there may be other factors you are not seeing in your assumptions that may or may not be see in a disassembly. there are obvious things that we would expect like one of the loop variables to be register based vs memory based. alignment, I wouldn't expect the clock settings to change if you used the same code, but a different set of experiments using timers or a scope you can measure the clock settings to see if they were configured the same. background tasks, interrupts and dumb luck as to how/when they hit the test. But bottom line, sometimes it is as simple as a missed optimization or subtle differences in how one compiler generates code to another, it is as often not those things and more of a system issue, memory speed, peripheral speed, caches or their architecture, how the various busses in the design operate, etc. For some of these cortex-ms (and many other processors) you can exploit their bus behavior to show a performance difference in something that the average person wouldn't expect to see.
Keil over-optimization (unlikely, because the code is very simple)
You cant over-optimize you can under/miss so if anything gcc missed something that Kiel didn't. Not the other way around
arm-none-eabi-gcc under-optimization due to wrong compiler flags (I use CLion Embedded plugins` CMakeLists.txt)
will see below but this is highly likely esp debug vs release, I never build for debug (never use a debugger) you have to test everything twice, and if you don't test as you go it makes it much harder to debug so the release version if it has issues takes a lot more work to figure out the issues.
A bug in the initialization so that chip has lower clock frequency with arm-none-eabi-gcc (to be investigated)
My guess is it isn't this, this would imply you made a really big mistake and didn't compile the same code on each tool so it wasn't a fair comparison.
Let's run it.
Using the systick timer, 24 bit (current value register address passed in r0)
.align 8
.thumb_func
.globl TEST
TEST:
push {r4,r5,r6,lr}
mov r4,r0
ldr r5,[r4]
bl soft_delay
ldr r3,[r4]
sub r0,r5,r3
pop {r4,r5,r6,pc}
to avoid overflowing the 24 bit timer the loops count to limited to 200000 times not 2000000 times. I assume the code you left out is 2000000 - 1. If not this still shows the relevant differences.
-O0 code
.align 8
soft_delay:
PUSH {r3,lr}
MOV r0,#0
STR r0,[sp,#0]
B L6.14
L6.8:
LDR r0,[sp,#0]
ADD r0,r0,#1
STR r0,[sp,#0]
L6.14:
LDR r1,L6.24
LDR r0,[sp,#0]
CMP r0,r1
BCC L6.8
POP {r3,pc}
.align
L6.24: .word 100000 - 1
08000200 <soft_delay>:
8000200: b508 push {r3, lr}
8000202: 2000 movs r0, #0
8000204: 9000 str r0, [sp, #0]
8000206: e002 b.n 800020e <L6.14>
08000208 <L6.8>:
8000208: 9800 ldr r0, [sp, #0]
800020a: 3001 adds r0, #1
800020c: 9000 str r0, [sp, #0]
0800020e <L6.14>:
800020e: 4902 ldr r1, [pc, #8] ; (8000218 <L6.24>)
8000210: 9800 ldr r0, [sp, #0]
8000212: 4288 cmp r0, r1
8000214: d3f8 bcc.n 8000208 <L6.8>
8000216: bd08 pop {r3, pc}
08000218 <L6.24>:
8000218: 0001869f
00124F8B systick timer ticks
-O1 code
soft_delay:
PUSH {r3,lr}
MOV r0,#0
STR r0,[sp,#0]
LDR r0,L6.24
B L6.16
L6.10:
LDR r1,[sp,#0]
ADD r1,r1,#1
STR r1,[sp,#0]
L6.16:
LDR r1,[sp,#0]
CMP r1,r0
BCC L6.10
POP {r3,pc}
.align
L6.24: .word 100000 - 1
08000200 <soft_delay>:
8000200: b508 push {r3, lr}
8000202: 2000 movs r0, #0
8000204: 9000 str r0, [sp, #0]
8000206: 4804 ldr r0, [pc, #16] ; (8000218 <L6.24>)
8000208: e002 b.n 8000210 <L6.16>
0800020a <L6.10>:
800020a: 9900 ldr r1, [sp, #0]
800020c: 3101 adds r1, #1
800020e: 9100 str r1, [sp, #0]
08000210 <L6.16>:
8000210: 9900 ldr r1, [sp, #0]
8000212: 4281 cmp r1, r0
8000214: d3f9 bcc.n 800020a <L6.10>
8000216: bd08 pop {r3, pc}
08000218 <L6.24>:
8000218: 0001869f
000F424E systicks
-O2 code
soft_delay:
SUB sp,sp,#4
MOVS r0,#0
STR r0,[sp,#0]
LDR r0,L4.24
L4.8:
LDR r1,[sp,#0]
ADDS r1,r1,#1
STR r1,[sp,#0]
CMP r1,r0
BCC L4.8
ADD sp,sp,#4
BX lr
.align
L4.24: .word 100000 - 1
08000200 <soft_delay>:
8000200: b081 sub sp, #4
8000202: 2000 movs r0, #0
8000204: 9000 str r0, [sp, #0]
8000206: 4804 ldr r0, [pc, #16] ; (8000218 <L4.24>)
08000208 <L4.8>:
8000208: 9900 ldr r1, [sp, #0]
800020a: 3101 adds r1, #1
800020c: 9100 str r1, [sp, #0]
800020e: 4281 cmp r1, r0
8000210: d3fa bcc.n 8000208 <L4.8>
8000212: b001 add sp, #4
8000214: 4770 bx lr
8000216: 46c0 nop ; (mov r8, r8)
08000218 <L4.24>:
8000218: 0001869f
000AAE65 systicks
-O3
soft_delay:
PUSH {r3,lr}
MOV r0,#0
STR r0,[sp,#0]
LDR r0,L5.20
L5.8:
LDR r1,[sp,#0]
ADD r1,r1,#1
STR r1,[sp,#0]
CMP r1,r0
BCC L5.8
POP {r3,pc}
.align
L5.20: .word 100000 - 1
08000200 <soft_delay>:
8000200: b508 push {r3, lr}
8000202: 2000 movs r0, #0
8000204: 9000 str r0, [sp, #0]
8000206: 4803 ldr r0, [pc, #12] ; (8000214 <L5.20>)
08000208 <L5.8>:
8000208: 9900 ldr r1, [sp, #0]
800020a: 3101 adds r1, #1
800020c: 9100 str r1, [sp, #0]
800020e: 4281 cmp r1, r0
8000210: d3fa bcc.n 8000208 <L5.8>
8000212: bd08 pop {r3, pc}
08000214 <L5.20>:
8000214: 0001869f
000AAE6A systicks
Interestingly alignment doesn't affect any of these results.
Comparing your results relative to each other and the above in a spreadsheet
18.7 1.000 00124F8B 1200011 1.000
13.3 0.711 000F424E 1000014 0.833
9.8 0.524 000AAE65 700005 0.583
9.9 0.529 000AAE6A 700010 0.583
It shows that the various stages as I have measured also show improvements and that -O3 is slightly slower.
Analyze what happened.
void soft_delay(void) {
for (volatile uint32_t i=0; i<2000000; ++i) { }
}
because this counts up AND is volatile the compiler cannot do the usual count down and save an instruction (subs then bne rather than add, cmp, bcc)
-O0 code
soft_delay:
PUSH {r3,lr} allocate space for i
MOV r0,#0 i = 0
STR r0,[sp,#0] i = 0
B L6.14
L6.8:
LDR r0,[sp,#0] read i from memory
ADD r0,r0,#1 increment i
STR r0,[sp,#0] save i to memory
L6.14:
LDR r1,L6.24 read max value
LDR r0,[sp,#0] read i from memory
CMP r0,r1 compare i and max value
BCC L6.8 branch if unsigned lower
POP {r3,pc} return
I should have examined the code first L6.24 should have been 2000000 not 2000000 - 1. You left this out of your question.
No optimization generally means just bang out the code in order as in the high level language.
r3 doesn't need to be preserved neither does LR but the variable is volatile so it needs space on the stack the compiler chose to do it this way for this optimization level pushing lr allows for it to pop pc at the end.
push is a pseudo instruction for stm (stmdb) so 8 is subtracted from the stack pointer then the registers are saved in order so if the sp was at 0x1008 then it changes to 0x1000 and writes r3 to 0x1000 and lr to 0x1004 so for the rest of this function it uses sp+0 which is 0x1000 in this example. The r3 and the push used in this way is to allocate a location for the variable i in the code.
-O1 version
soft_delay:
PUSH {r3,lr} allocate space
MOV r0,#0 i = 0
STR r0,[sp,#0] i = 0
LDR r0,L6.24 read max/test value
B L6.16
L6.10:
LDR r1,[sp,#0] load i from memory
ADD r1,r1,#1 increment i
STR r1,[sp,#0] save i to memory
L6.16:
LDR r1,[sp,#0] read i from memory
CMP r1,r0 compare i with test value
BCC L6.10 branch if unsigned lower
POP {r3,pc}
The primary difference between -O0 and -O1 in this case is the -O0 version reads the max value every time through the loop. The -O1 version reads it outside the loop one time.
-O0
08000208 <L6.8>:
8000208: 9800 ldr r0, [sp, #0]
800020a: 3001 adds r0, #1
800020c: 9000 str r0, [sp, #0]
800020e: 4902 ldr r1, [pc, #8] ; (8000218 <L6.24>)
8000210: 9800 ldr r0, [sp, #0]
8000212: 4288 cmp r0, r1
8000214: d3f8 bcc.n 8000208 <L6.8>
1200011 / 100000 = 12
The bulk of the time is in the above loop. 7 instructions three loads two stores. That is 12 things so perhaps its one clock per.
-O1 code
0800020a <L6.10>:
800020a: 9900 ldr r1, [sp, #0]
800020c: 3101 adds r1, #1
800020e: 9100 str r1, [sp, #0]
08000210 <L6.16>:
8000210: 9900 ldr r1, [sp, #0]
8000212: 4281 cmp r1, r0
8000214: d3f9 bcc.n 800020a <L6.10>
1000014 / 100000 = 10
0800020a <L6.10>:
800020a: 9900 ldr r1, [sp, #0]
800020c: 3101 adds r1, #1
800020e: 9100 str r1, [sp, #0]
8000210: 9900 ldr r1, [sp, #0]
8000212: 4281 cmp r1, r0
8000214: d3f9 bcc.n 800020a <L6.10>
6 instructions, two loads one store. 8 things 10 clocks. The difference here from -O0 is that the compare value is read before/outside the loop so that saves that instruction and that memory cycle.
-O2 code
08000208 <L4.8>:
8000208: 9900 ldr r1, [sp, #0]
800020a: 3101 adds r1, #1
800020c: 9100 str r1, [sp, #0]
800020e: 4281 cmp r1, r0
8000210: d3fa bcc.n 8000208 <L4.8>
700005 / 100000 = 7 ticks per loop
So by some folks definition, this isn't honoring the volatile, or is it? The compare value is outside the loop and the way this is written it should be 2000000 + 1, yes? It reads i from memory one time per loop rather than twice but does store it every time through the loop with the new value. Basically it removed the second load and that saved some time waiting on that read to finish.
-O3 code
08000208 <L5.8>:
8000208: 9900 ldr r1, [sp, #0]
800020a: 3101 adds r1, #1
800020c: 9100 str r1, [sp, #0]
800020e: 4281 cmp r1, r0
8000210: d3fa bcc.n 8000208 <L5.8>
The inner loop is the same as -O2.
-O2 does this
08000200 <soft_delay>:
8000200: b081 sub sp, #4
8000202: 2000 movs r0, #0
8000204: 9000 str r0, [sp, #0]
8000206: 4804 ldr r0, [pc, #16] ; (8000218 <L4.24>)
...
8000212: b001 add sp, #4
8000214: 4770 bx lr
-O3 does this
08000200 <soft_delay>:
8000200: b508 push {r3, lr}
8000202: 2000 movs r0, #0
8000204: 9000 str r0, [sp, #0]
8000206: 4803 ldr r0, [pc, #12] ; (8000214 <L5.20>)
8000212: bd08 pop {r3, pc}
Now that is fewer instructions yes, but the push and pop take longer they have memory cycle overhead, the subtract and add of the stack pointer instructions are faster than those memory cycles even with the fewer instructions. So the subtle difference in time is the push/pop outside the loop.
Now for GCC (9.2.0)
For starters I don't know if Kiel was targetted at thumb in general (all variants) the cortex-ms or the cortex-m3 specifically.
First -O0 code:
-O0
soft_delay:
push {r7, lr}
sub sp, sp, #8
add r7, sp, #0
movs r3, #0
str r3, [r7, #4]
b .L2
.L3:
ldr r3, [r7, #4]
adds r3, r3, #1
str r3, [r7, #4]
.L2:
ldr r3, [r7, #4]
ldr r2, .L4
cmp r3, r2
bls .L3
nop
nop
mov sp, r7
add sp, sp, #8
# sp needed
pop {r7}
pop {r0}
bx r0
.L5:
.align 2
.L4:
.word 199999
08000200 <soft_delay>:
8000200: b580 push {r7, lr}
8000202: b082 sub sp, #8
8000204: af00 add r7, sp, #0
8000206: 2300 movs r3, #0
8000208: 607b str r3, [r7, #4]
800020a: e002 b.n 8000212 <soft_delay+0x12>
800020c: 687b ldr r3, [r7, #4]
800020e: 3301 adds r3, #1
8000210: 607b str r3, [r7, #4]
8000212: 687b ldr r3, [r7, #4]
8000214: 4a04 ldr r2, [pc, #16] ; (8000228 <soft_delay+0x28>)
8000216: 4293 cmp r3, r2
8000218: d9f8 bls.n 800020c <soft_delay+0xc>
800021a: 46c0 nop ; (mov r8, r8)
800021c: 46c0 nop ; (mov r8, r8)
800021e: 46bd mov sp, r7
8000220: b002 add sp, #8
8000222: bc80 pop {r7}
8000224: bc01 pop {r0}
8000226: 4700 bx r0
8000228: 00030d3f andeq r0, r3, pc, lsr sp
00124F9F
Immediately we see two things, first the stack frame which Kiel was not building and second these mystery nops after the compare, gotta be some chip errata or something, need to look that up. From my other answer that may be deleted by now gcc 5.4.0 put one nop, tcc 9.2.0 put two. so this loop has
1200031 / 100000 = 12 ticks per loop
800020c: 687b ldr r3, [r7, #4]
800020e: 3301 adds r3, #1
8000210: 607b str r3, [r7, #4]
8000212: 687b ldr r3, [r7, #4]
8000214: 4a04 ldr r2, [pc, #16] ; (8000228 <soft_delay+0x28>)
8000216: 4293 cmp r3, r2
8000218: d9f8 bls.n 800020c <soft_delay+0xc>
The main loop where this code spends its time is also 12 ticks like Kiel its the same just different registers which don't matter. The subtle overall time difference is that the stack frame and the extra nops make the gcc version slightly longer.
arm-none-eabi-gcc -O0 -fomit-frame-pointer -c -mthumb -mcpu=cortex-m0 hello.c -o hello.o
arm-none-eabi-objdump -D hello.o > hello.list
arm-none-eabi-gcc -O0 -fomit-frame-pointer -S -mthumb -mcpu=cortex-m0 hello.c
If I build without a frame pointer then gcc -O0 becomes
soft_delay:
sub sp, sp, #8
movs r3, #0
str r3, [sp, #4]
b .L2
.L3:
ldr r3, [sp, #4]
adds r3, r3, #1
str r3, [sp, #4]
.L2:
ldr r3, [sp, #4]
ldr r2, .L4
cmp r3, r2
bls .L3
nop
nop
add sp, sp, #8
bx lr
.L5:
.align 2
.L4:
.word 99999
08000200 <soft_delay>:
8000200: b082 sub sp, #8
8000202: 2300 movs r3, #0
8000204: 9301 str r3, [sp, #4]
8000206: e002 b.n 800020e <soft_delay+0xe>
8000208: 9b01 ldr r3, [sp, #4]
800020a: 3301 adds r3, #1
800020c: 9301 str r3, [sp, #4]
800020e: 9b01 ldr r3, [sp, #4]
8000210: 4a03 ldr r2, [pc, #12] ; (8000220 <soft_delay+0x20>)
8000212: 4293 cmp r3, r2
8000214: d9f8 bls.n 8000208 <soft_delay+0x8>
8000216: 46c0 nop ; (mov r8, r8)
8000218: 46c0 nop ; (mov r8, r8)
800021a: b002 add sp, #8
800021c: 4770 bx lr
800021e: 46c0 nop ; (mov r8, r8)
8000220: 0001869f
00124F94
and saves 11 clocks over the other gcc version unlike Kiel gcc is not doing the push pop thing so saving some clocks over Kiel but the nops don't help.
Update: I had the wrong number of loops for Kiel because it used unsigned lower instead of unsigned lower or same as with gcc. Even the playing field, remove the nops fix the loops gcc is 00124F92 and Kiel 00124F97 5 clocks slower due to the push/pop vs sp math. gcc 5.4.0 also does the sp math thing, with the nop 00124F93. Being outside the loop stuff these differences while measurable are also in the noise when comparing these two (three) compilers.
gcc -O1
soft_delay:
sub sp, sp, #8
mov r3, #0
str r3, [sp, #4]
ldr r2, [sp, #4]
ldr r3, .L5
cmp r2, r3
bhi .L1
mov r2, r3
.L3:
ldr r3, [sp, #4]
add r3, r3, #1
str r3, [sp, #4]
ldr r3, [sp, #4]
cmp r3, r2
bls .L3
.L1:
add sp, sp, #8
bx lr
.L6:
.align 2
.L5:
.word 99999
08000200 <soft_delay>:
8000200: b082 sub sp, #8
8000202: 2300 movs r3, #0
8000204: 9301 str r3, [sp, #4]
8000206: 9a01 ldr r2, [sp, #4]
8000208: 4b05 ldr r3, [pc, #20] ; (8000220 <soft_delay+0x20>)
800020a: 429a cmp r2, r3
800020c: d806 bhi.n 800021c <soft_delay+0x1c>
800020e: 1c1a adds r2, r3, #0
8000210: 9b01 ldr r3, [sp, #4]
8000212: 3301 adds r3, #1
8000214: 9301 str r3, [sp, #4]
8000216: 9b01 ldr r3, [sp, #4]
8000218: 4293 cmp r3, r2
800021a: d9f9 bls.n 8000210 <soft_delay+0x10>
800021c: b002 add sp, #8
800021e: 4770 bx lr
8000220: 0001869f muleq r1, pc, r6 ; <UNPREDICTABLE>
000F4251
10 ticks per loop
8000210: 9b01 ldr r3, [sp, #4]
8000212: 3301 adds r3, #1
8000214: 9301 str r3, [sp, #4]
8000216: 9b01 ldr r3, [sp, #4]
8000218: 4293 cmp r3, r2
800021a: d9f9 bls.n 8000210 <soft_delay+0x10>
Same as Kiel the load of the compare value is outside the loop now saving a little per loop. It was architected a little different. And I believe the nops after the bls are something else. I just saw someone asking about why gcc did something that another didn't what seemed to be an extra instruction. I would use the term missed optimization vs bug, but either way this one doesn't have the nops...
gcc -O2 code
soft_delay:
mov r3, #0
sub sp, sp, #8
str r3, [sp, #4]
ldr r3, [sp, #4]
ldr r2, .L7
cmp r3, r2
bhi .L1
.L3:
ldr r3, [sp, #4]
add r3, r3, #1
str r3, [sp, #4]
ldr r3, [sp, #4]
cmp r3, r2
bls .L3
.L1:
add sp, sp, #8
bx lr
.L8:
.align 2
.L7:
.word 99999
08000200 <soft_delay>:
8000200: 2300 movs r3, #0
8000202: b082 sub sp, #8
8000204: 9301 str r3, [sp, #4]
8000206: 9b01 ldr r3, [sp, #4]
8000208: 4a05 ldr r2, [pc, #20] ; (8000220 <soft_delay+0x20>)
800020a: 4293 cmp r3, r2
800020c: d805 bhi.n 800021a <soft_delay+0x1a>
800020e: 9b01 ldr r3, [sp, #4]
8000210: 3301 adds r3, #1
8000212: 9301 str r3, [sp, #4]
8000214: 9b01 ldr r3, [sp, #4]
8000216: 4293 cmp r3, r2
8000218: d9f9 bls.n 800020e <soft_delay+0xe>
800021a: b002 add sp, #8
800021c: 4770 bx lr
800021e: 46c0 nop ; (mov r8, r8)
8000220: 0001869f
000F4251
No difference from -O1
800020e: 9b01 ldr r3, [sp, #4]
8000210: 3301 adds r3, #1
8000212: 9301 str r3, [sp, #4]
8000214: 9b01 ldr r3, [sp, #4]
8000216: 4293 cmp r3, r2
8000218: d9f9 bls.n 800020e <soft_delay+0xe>
gcc is not willing to take that second load out of the loop.
at the -O2 level Kiel is 70005 ticks and gcc 1000017. 42 percent more/slower.
gcc -O3 produced the same code as -O2.
So the key difference here is perhaps an interpretation of what volatile does, and there are some folks at SO that get upset about its use anyway, but let's just assume that it means everything you do with the variable needs to go to/from memory.
From what I normally see that means this
.L3:
ldr r3, [sp, #4]
add r3, r3, #1
str r3, [sp, #4]
ldr r3, [sp, #4]
cmp r3, r2
bls .L3
not this
.L3:
ldr r3, [sp, #4]
add r3, r3, #1
str r3, [sp, #4]
cmp r3, r2
bls .L3
Is that a Kiel bug? Do you want to use your over-optimization term here?
There are two operations an increment
ldr r3, [sp, #4]
add r3, r3, #1
str r3, [sp, #4]
and a compare
ldr r3, [sp, #4]
cmp r3, r2
bls .L3
arguably each should access the variable from memory not from a register. (in a pure debug version sense you should see code like this too btw, although the tool defines what it means by debug version)
When you figure out which gcc you have and how it was used it may account for even more code on the gcc side being the 100% slower not 40%.
I don't know that you could make this any tighter, I don't think re-arranging instructions will improve performance either.
Also, this was a missed optimization in gcc:
cmp r3, r2
bhi .L1
gcc knew that it was starting from zero and knew it was going to a bigger number so r3 would never be larger than r2 here.
We wish for the tool to make this:
soft_delay:
mov r3, #0
ldr r2, .L7
.L3:
add r3, r3, #1
cmp r3, r2
bls .L3
.L1:
bx lr
.L8:
.align 2
.L7:
.word 99999
00061A88
at 4 instructions per loop on average
but without the volatile it is dead code so the optimizer would simply remove it rather than make this code. A down count loop would be slightly smaller
soft_delay:
ldr r2, .L7
.L3:
sub r2, r2, #1
bne .L3
.L1:
bx lr
.L8:
.align 2
.L7:
.word 100000
000493E7
3 ticks per loop, removing the extra instruction helped.
Keil over-optimization (unlikely, because the code is very simple)
You might actually be right here, not because it is simple, but what does volatile really mean, and is it subject to interpretation by the compilers (I would have to find a spec). Is this a Kiel bug, did it over optimize?
There still isn't such a thing as over-optimization, there is a name for that, a compiler bug. So did Kiel interpret this wrong or Kiel and gcc disagree on the interpretation of volatile.
arm-none-eabi-gcc under-optimization due to wrong compiler flags (I use CLion Embedded plugins` CMakeLists.txt)
This could be it as well, for the same reason. Is this simply an "implementation defined" difference between compilers and both are right based on their definition?
Now gcc did miss an optimization here (or two), but it accounts for a small amount as it is outside the loop.
GCC ||| KEIL
|||
soft_delay: |||
mov r3, #0 |||
sub sp, sp, #8 |||
str r3, [sp, #4] |||
ldr r3, [sp, #4] |||
ldr r2, .L7 |||
cmp r3, r2 |||
bhi .L1 ||| soft_delay PROC
.L3: ||| PUSH {r3,lr}
ldr r3, [sp, #4] ||| MOVS r0,#0
add r3, r3, #1 ||| STR r0,[sp,#0]
str r3, [sp, #4] ||| LDR r0,|L5.20|
ldr r3, [sp, #4] ||| |L5.8|
cmp r3, r2 ||| LDR r1,[sp,#0]
bls .L3 ||| ADDS r1,r1,#1
.L1: ||| STR r1,[sp,#0]
add sp, sp, #8 ||| CMP r1,r0
bx lr ||| BCC |L5.8|
.L7: ||| POP {r3,pc}
.word 1999999 ||| ENDP
There is obvious bug in KEIL. volatile means that its value has to be loaded before every use and saved when changed. ? Keil is missing one load.
The variable is used 2 times: 1: when increased, 2: when compared. Two loads needed.

ARM Disassembly - confused about "LDR r7, [pc, #0x14]"

I am attempting to learn ARM assembly. I decided to disassembly the "read" function and this is what I get. From the looks of it, it seems to be making a system call (svc #0) using the R7 register as the system call number.
mov ip, r7 # save R7
ldr r7, [pc, #0x14] # get system call number and put it into R7 ??
svc #0 # make system call
mov r7, ip # restore R7
cmn r0, #0x1000
bxls lr
rsb r0, r0, #0 # R0 = 0
b #2976848216
I am a bit confused though on why it is loading the system call number the way it is ("LDR r7, [PC, #0x14]"). Isn't this just doing in C code r7 = *(pc + 0x14)? I looked at other functions that might also use system calls (e.g. kill, wait, etc.) and they use a very similar convention (i.e. LDR R7, [PC, #0x14]).
This is on Android if it helps at all.
Thanks!
mov ip, r7 ## save R7
ldr r7, [pc, #0x14] ## get system call number and put it into R7 ??
svc #0 ## make system call
mov r7, ip ## restore R7
cmn r0, #0x1000 #
bxls lr #
rsb r0, r0, #0 ## R0 = 0
.word 0x1234
.word 0xABCD
you pretty much left out the most important parts so had to improvise
00000000 <.text>:
0: e1a0c007 mov ip, r7
4: e59f7014 ldr r7, [pc, #20] ; 20 <.text+0x20>
8: ef000000 svc 0x00000000
c: e1a0700c mov r7, ip
10: e3700a01 cmn r0, #4096 ; 0x1000
14: 912fff1e bxls lr
18: e2600000 rsb r0, r0, #0
1c: 00001234 andeq r1, r0, r4, lsr r2
20: 0000abcd andeq sl, r0, sp, asr #23
And yes it is doing what you say it is doing, it is loading some value in r7 before making the system call, now what value is it as to why it is using a pc relative load (likely a constant that wont fit as an immediate, and/or a link time resolved value rather than compile time) and are there different values for different system calls and is r7 a parameter or not? Well you didnt provide enough information to talk about that. Once you have/see that information then that should be pretty obvious what those answers are...if any of those are is your question.

ARM Cortex-M3 startup file

I am modifying a startup file for an ARM Cortex-M3 microcontroller. Everything works fine so far, but I have a question regarding the need of using assembler code to perform the zero-filling of the BSS block.
By default the reset interrupt in the startup file looks as follows:
// Zero fill the bss segment.
__asm( " ldr r0, =_bss\n"
" ldr r1, =_ebss\n"
" mov r2, #0\n"
" .thumb_func\n"
" zero_loop:\n"
" cmp r0, r1\n"
" it lt\n"
" strlt r2, [r0], #4\n"
" blt zero_loop"
);
Using that code everything works as expected. However if I change the previous code for the following it stops working:
// Zero fill the bss segment.
for(pui32Dest = &_bss; pui32Dest < &_ebss; )
{
*pui32Dest++ = 0;
}
In principle both codes should do the same (fill the BSS with zeros), but the second one does not work for some reason that I fail to understand. I belive that the .thumb_func directive must play a role here, but I am not very familiar with ARM assembler. Any ideas or directions to help me understanding? Thanks!
Edit: By the way, the code to initialize the data segment (e.g. copy from Flash to RAM) is as follows and works just fine.
// Copy the data segment initializers from flash to SRAM.
pui32Src = &_etext;
for(pui32Dest = &_data; pui32Dest < &_edata; )
{
*pui32Dest++ = *pui32Src++;
}
Edit: Added the dissasembled code for both functions.
Assembly for the first looks like:
2003bc: 4806 ldr r0, [pc, #24] ; (2003d8 <zero_loop+0x14>)
2003be: 4907 ldr r1, [pc, #28] ; (2003dc <zero_loop+0x18>)
2003c0: f04f 0200 mov.w r2, #0
002003c4 <zero_loop>:
2003c4: 4288 cmp r0, r1
2003c6: bfb8 it lt
2003c8: f840 2b04 strlt.w r2, [r0], #4
2003cc: dbfa blt.n 2003c4 <zero_loop>
Assembly for the second looks like:
2003bc: f645 5318 movw r3, #23832 ; 0x5d18
2003c0: f2c2 0300 movt r3, #8192 ; 0x2000
2003c4: 9300 str r3, [sp, #0]
2003c6: e004 b.n 2003d2 <ResetISR+0x6e>
2003c8: 9b00 ldr r3, [sp, #0]
2003ca: 1d1a adds r2, r3, #4
2003cc: 9200 str r2, [sp, #0]
2003ce: 2200 movs r2, #0
2003d0: 601a str r2, [r3, #0]
2003d2: 9a00 ldr r2, [sp, #0]
2003d4: f644 033c movw r3, #18492 ; 0x483c
2003d8: f2c2 0300 movt r3, #8192 ; 0x2000
2003dc: 429a cmp r2, r3
2003de: d3f3 bcc.n 2003c8 <ResetISR+0x64>
If the initial stack is in the .bss section as suggested, you can see from the disassembly why the C code fails - it's loading the current pointer from the stack, saving the incremented pointer back to the stack, zeroing the location, then reloading the incremented pointer for the next iteration. If you zero the contents of the stack while using them, Bad Things happen.
In this case, turning on optimisation might fix it (a smart compiler should generate pretty much the same as the assembly code if it actually tries). More generally, though, it's probably safer to consider sticking with assembly code when doing things like this that would normally be done at a level below the C runtime environment - bootstrapping a C environment from C code which expects that environment to exist already is risky at best, since you can only hope the code doesn't attempt to use anything that's not yet set up.
After a quick look around (I'm not overly familiar with the specifics of Cortex-M development), it seems an alternative/additional solution might be adjusting the linker script to move the stack somewhere else.

Local variable location in memory

For a homework assignment I have been given some c files, and compiled them using arm-linux-gcc (we will eventually be targeting gumstix boards, but for these exercises we have been working with qemu and ema).
One of the questions confuses me a bit-- we are told to:
Use arm-linux-objdump to find the location of variables declared in main() in the executable binary.
However, these variables are local and thus shouldn't have addresses until runtime, correct?
I'm thinking that maybe what I need to find is the offset in the stack frame, which can in fact be found using objdump (not that I know how).
Anyways, any insight into the matter would be greatly appreciated, and I would be happy to post the source code if necessary.
unsigned int one ( unsigned int, unsigned int );
unsigned int two ( unsigned int, unsigned int );
unsigned int myfun ( unsigned int x, unsigned int y, unsigned int z )
{
unsigned int a,b;
a=one(x,y);
b=two(a,z);
return(a+b);
}
compile and disassemble
arm-none-eabi-gcc -c fun.c -o fun.o
arm-none-eabi-objdump -D fun.o
code created by compiler
00000000 <myfun>:
0: e92d4800 push {fp, lr}
4: e28db004 add fp, sp, #4
8: e24dd018 sub sp, sp, #24
c: e50b0010 str r0, [fp, #-16]
10: e50b1014 str r1, [fp, #-20]
14: e50b2018 str r2, [fp, #-24]
18: e51b0010 ldr r0, [fp, #-16]
1c: e51b1014 ldr r1, [fp, #-20]
20: ebfffffe bl 0 <one>
24: e50b0008 str r0, [fp, #-8]
28: e51b0008 ldr r0, [fp, #-8]
2c: e51b1018 ldr r1, [fp, #-24]
30: ebfffffe bl 0 <two>
34: e50b000c str r0, [fp, #-12]
38: e51b2008 ldr r2, [fp, #-8]
3c: e51b300c ldr r3, [fp, #-12]
40: e0823003 add r3, r2, r3
44: e1a00003 mov r0, r3
48: e24bd004 sub sp, fp, #4
4c: e8bd4800 pop {fp, lr}
50: e12fff1e bx lr
Short answer is the memory is "allocated" both at compile time and at run time. At compile time in the sense that the compiler at compile time determines the size of the stack frame and who goes where. Run time in the sense that the memory itself is on the stack which is a dynamic thing. The stack frame is taken from stack memory at run time, almost like a malloc() and free().
It helps to know the calling convention, x enters in r0, y in r1, z in r2. then x has its home at fp-16, y at fp-20, and z at fp-24. then the call to one() needs x and y so it pulls those from the stack (x and y). the result of one() goes into a which is saved at fp-8 so that is the home for a. and so on.
the function one is not really at address 0, this is a disassembly of an object file not a linked binary. once an object is linked in with the rest of the objects and libraries, the missing parts, like where external functions are, are patched in by the linker and the calls to one() and two() will get real addresses. (and the program will likely not start at address 0).
I cheated here a little, I knew that with no optimizations enabled on the compiler and a relatively simple function like this there really is no reason for a stack frame:
compile with just a little optimization
arm-none-eabi-gcc -O1 -c fun.c -o fun.o
arm-none-eabi-objdump -D fun.o
and the stack frame is gone, the local variables remain in registers.
00000000 :
0: e92d4038 push {r3, r4, r5, lr}
4: e1a05002 mov r5, r2
8: ebfffffe bl 0
c: e1a04000 mov r4, r0
10: e1a01005 mov r1, r5
14: ebfffffe bl 0
18: e0800004 add r0, r0, r4
1c: e8bd4038 pop {r3, r4, r5, lr}
20: e12fff1e bx lr
what the compiler decided to do instead is give itself more registers to work with by saving them on the stack. Why it saved r3 is a mystery, but that is another topic...
entering the function r0 = x, r1 = y and r2 = z per the calling convention, we can leave r0 and r1 alone (try again with one(y,x) and see what happens) since they drop right into one() and are never used again. The calling convention says that r0-r3 can be destroyed by a function, so we need to preserve z for later so we save it in r5. The result of one() is r0 per the calling convention, since two() can destroy r0-r3 we need to save a for later, after the call to two() also we need r0 for the call to two anyway, so r4 now holds a. We saved z in r5 (was in r2 moved to r5) before the call to one, we need the result of one() as the first parameter to two(), and it is already there, we need z as the second so we move r5 where we had saved z to r1, then we call two(). the result of two() per the calling convention. Since b + a = a + b from basic math properties the final add before returning is r0 + r4 which is b + a, and the result goes in r0 which is the register used to return something from a function, per the convention. clean up the stack and restore the modified registers, done.
Since myfun() made calls to other functions using bl, bl modifies the link register (r14), in order to be able to return from myfun() we need the value in the link register to be preserved from the entry into the function to the final return (bx lr), so lr is pushed on the stack. The convention states that we can destroy r0-r3 in our function but not other registers so r4 and r5 are pushed on the stack because we used them. why r3 is pushed on the stack is not necessary from a calling convention perspective, I wonder if it was done in anticipation of a 64 bit memory system, making two full 64 bit writes is cheaper than one 64 bit write and one 32 bit right. but you would need to know the alignment of the stack going in so that is just a theory. There is no reason to preserve r3 in this code.
Now take this knowledge and disassemble the code assigned (arm-...-objdump -D something.something) and do the same kind of analysis. particularly with functions named main() vs functions not named main (I did not use main() on purpose) the stack frame can be a size that doesnt make sense, or less sense than other functions. In the non optimized case above we needed to store 6 things total, x,y,z,a,b and the link register 6*4 = 24 bytes which resulted in sub sp, sp, #24, I need to think about the stack pointer vs frame pointer
thing for a bit. I think there is a command line argument to tell the compiler not to use a frame pointer. -fomit-frame-pointer and it saves a couple of instructions
00000000 <myfun>:
0: e52de004 push {lr} ; (str lr, [sp, #-4]!)
4: e24dd01c sub sp, sp, #28
8: e58d000c str r0, [sp, #12]
c: e58d1008 str r1, [sp, #8]
10: e58d2004 str r2, [sp, #4]
14: e59d000c ldr r0, [sp, #12]
18: e59d1008 ldr r1, [sp, #8]
1c: ebfffffe bl 0 <one>
20: e58d0014 str r0, [sp, #20]
24: e59d0014 ldr r0, [sp, #20]
28: e59d1004 ldr r1, [sp, #4]
2c: ebfffffe bl 0 <two>
30: e58d0010 str r0, [sp, #16]
34: e59d2014 ldr r2, [sp, #20]
38: e59d3010 ldr r3, [sp, #16]
3c: e0823003 add r3, r2, r3
40: e1a00003 mov r0, r3
44: e28dd01c add sp, sp, #28
48: e49de004 pop {lr} ; (ldr lr, [sp], #4)
4c: e12fff1e bx lr
optimizing saves a whole lot more though...
It's going to depend on the program and how exactly they want the location of the variables. Does the question want what code section they're stored in? .const .bss etc? Does it want specific addresses? Either way a good start is using objdump -S flag
objdump -S myprogram > dump.txt
This is nice because it will print out an intermixing of your source code and the assembly with addresses. From here just do a search for your int main and that should get you started.

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