Why is gcc putting NOPs at the end of functions? [duplicate] - c

I've been working with C for a short while and very recently started to get into ASM. When I compile a program:
int main(void)
{
int a = 0;
a += 1;
return 0;
}
The objdump disassembly has the code, but nops after the ret:
...
08048394 <main>:
8048394: 55 push %ebp
8048395: 89 e5 mov %esp,%ebp
8048397: 83 ec 10 sub $0x10,%esp
804839a: c7 45 fc 00 00 00 00 movl $0x0,-0x4(%ebp)
80483a1: 83 45 fc 01 addl $0x1,-0x4(%ebp)
80483a5: b8 00 00 00 00 mov $0x0,%eax
80483aa: c9 leave
80483ab: c3 ret
80483ac: 90 nop
80483ad: 90 nop
80483ae: 90 nop
80483af: 90 nop
...
From what I learned nops do nothing, and since after ret wouldn't even be executed.
My question is: why bother? Couldn't ELF(linux-x86) work with a .text section(+main) of any size?
I'd appreciate any help, just trying to learn.

First of all, gcc doesn't always do this. The padding is controlled by -falign-functions, which is automatically turned on by -O2 and -O3:
-falign-functions
-falign-functions=n
Align the start of functions to the next power-of-two greater than n, skipping up to n bytes. For instance,
-falign-functions=32 aligns functions to the next 32-byte boundary, but -falign-functions=24 would align to the next 32-byte boundary only
if this can be done by skipping 23 bytes or less.
-fno-align-functions and -falign-functions=1 are equivalent and mean that functions will not be aligned.
Some assemblers only support this flag when n is a power of two; in
that case, it is rounded up.
If n is not specified or is zero, use a machine-dependent default.
Enabled at levels -O2, -O3.
There could be multiple reasons for doing this, but the main one on x86 is probably this:
Most processors fetch instructions in aligned 16-byte or 32-byte blocks. It can be
advantageous to align critical loop entries and subroutine entries by 16 in order to minimize
the number of 16-byte boundaries in the code. Alternatively, make sure that there is no 16-byte boundary in the first few instructions after a critical loop entry or subroutine entry.
(Quoted from "Optimizing subroutines in assembly
language" by Agner Fog.)
edit: Here is an example that demonstrates the padding:
// align.c
int f(void) { return 0; }
int g(void) { return 0; }
When compiled using gcc 4.4.5 with default settings, I get:
align.o: file format elf64-x86-64
Disassembly of section .text:
0000000000000000 <f>:
0: 55 push %rbp
1: 48 89 e5 mov %rsp,%rbp
4: b8 00 00 00 00 mov $0x0,%eax
9: c9 leaveq
a: c3 retq
000000000000000b <g>:
b: 55 push %rbp
c: 48 89 e5 mov %rsp,%rbp
f: b8 00 00 00 00 mov $0x0,%eax
14: c9 leaveq
15: c3 retq
Specifying -falign-functions gives:
align.o: file format elf64-x86-64
Disassembly of section .text:
0000000000000000 <f>:
0: 55 push %rbp
1: 48 89 e5 mov %rsp,%rbp
4: b8 00 00 00 00 mov $0x0,%eax
9: c9 leaveq
a: c3 retq
b: eb 03 jmp 10 <g>
d: 90 nop
e: 90 nop
f: 90 nop
0000000000000010 <g>:
10: 55 push %rbp
11: 48 89 e5 mov %rsp,%rbp
14: b8 00 00 00 00 mov $0x0,%eax
19: c9 leaveq
1a: c3 retq

This is done to align the next function by 8, 16 or 32-byte boundary.
From “Optimizing subroutines in assembly language” by A.Fog:
11.5 Alignment of code
Most microprocessors fetch code in aligned 16-byte or 32-byte blocks. If an importantsubroutine entry or jump label happens to be near the end of a 16-byte block then themicroprocessor will only get a few useful bytes of code when fetching that block of code. Itmay have to fetch the next 16 bytes too before it can decode the first instructions after thelabel. This can be avoided by aligning important subroutine entries and loop entries by 16.
[...]
Aligning a subroutine entry is as simple as putting as many
NOP
's as needed before thesubroutine entry to make the address divisible by 8, 16, 32 or 64, as desired.

As far as I remember, instructions are pipelined in cpu and different cpu blocks (loader, decoder and such) process subsequent instructions. When RET instructions is being executed, few next instructions are already loaded into cpu pipeline. It's a guess, but you can start digging here and if you find out (maybe the specific number of NOPs that are safe, share your findings please.

Related

Why are there so many nops at the end of the method? [duplicate]

I've been working with C for a short while and very recently started to get into ASM. When I compile a program:
int main(void)
{
int a = 0;
a += 1;
return 0;
}
The objdump disassembly has the code, but nops after the ret:
...
08048394 <main>:
8048394: 55 push %ebp
8048395: 89 e5 mov %esp,%ebp
8048397: 83 ec 10 sub $0x10,%esp
804839a: c7 45 fc 00 00 00 00 movl $0x0,-0x4(%ebp)
80483a1: 83 45 fc 01 addl $0x1,-0x4(%ebp)
80483a5: b8 00 00 00 00 mov $0x0,%eax
80483aa: c9 leave
80483ab: c3 ret
80483ac: 90 nop
80483ad: 90 nop
80483ae: 90 nop
80483af: 90 nop
...
From what I learned nops do nothing, and since after ret wouldn't even be executed.
My question is: why bother? Couldn't ELF(linux-x86) work with a .text section(+main) of any size?
I'd appreciate any help, just trying to learn.
First of all, gcc doesn't always do this. The padding is controlled by -falign-functions, which is automatically turned on by -O2 and -O3:
-falign-functions
-falign-functions=n
Align the start of functions to the next power-of-two greater than n, skipping up to n bytes. For instance,
-falign-functions=32 aligns functions to the next 32-byte boundary, but -falign-functions=24 would align to the next 32-byte boundary only
if this can be done by skipping 23 bytes or less.
-fno-align-functions and -falign-functions=1 are equivalent and mean that functions will not be aligned.
Some assemblers only support this flag when n is a power of two; in
that case, it is rounded up.
If n is not specified or is zero, use a machine-dependent default.
Enabled at levels -O2, -O3.
There could be multiple reasons for doing this, but the main one on x86 is probably this:
Most processors fetch instructions in aligned 16-byte or 32-byte blocks. It can be
advantageous to align critical loop entries and subroutine entries by 16 in order to minimize
the number of 16-byte boundaries in the code. Alternatively, make sure that there is no 16-byte boundary in the first few instructions after a critical loop entry or subroutine entry.
(Quoted from "Optimizing subroutines in assembly
language" by Agner Fog.)
edit: Here is an example that demonstrates the padding:
// align.c
int f(void) { return 0; }
int g(void) { return 0; }
When compiled using gcc 4.4.5 with default settings, I get:
align.o: file format elf64-x86-64
Disassembly of section .text:
0000000000000000 <f>:
0: 55 push %rbp
1: 48 89 e5 mov %rsp,%rbp
4: b8 00 00 00 00 mov $0x0,%eax
9: c9 leaveq
a: c3 retq
000000000000000b <g>:
b: 55 push %rbp
c: 48 89 e5 mov %rsp,%rbp
f: b8 00 00 00 00 mov $0x0,%eax
14: c9 leaveq
15: c3 retq
Specifying -falign-functions gives:
align.o: file format elf64-x86-64
Disassembly of section .text:
0000000000000000 <f>:
0: 55 push %rbp
1: 48 89 e5 mov %rsp,%rbp
4: b8 00 00 00 00 mov $0x0,%eax
9: c9 leaveq
a: c3 retq
b: eb 03 jmp 10 <g>
d: 90 nop
e: 90 nop
f: 90 nop
0000000000000010 <g>:
10: 55 push %rbp
11: 48 89 e5 mov %rsp,%rbp
14: b8 00 00 00 00 mov $0x0,%eax
19: c9 leaveq
1a: c3 retq
This is done to align the next function by 8, 16 or 32-byte boundary.
From “Optimizing subroutines in assembly language” by A.Fog:
11.5 Alignment of code
Most microprocessors fetch code in aligned 16-byte or 32-byte blocks. If an importantsubroutine entry or jump label happens to be near the end of a 16-byte block then themicroprocessor will only get a few useful bytes of code when fetching that block of code. Itmay have to fetch the next 16 bytes too before it can decode the first instructions after thelabel. This can be avoided by aligning important subroutine entries and loop entries by 16.
[...]
Aligning a subroutine entry is as simple as putting as many
NOP
's as needed before thesubroutine entry to make the address divisible by 8, 16, 32 or 64, as desired.
As far as I remember, instructions are pipelined in cpu and different cpu blocks (loader, decoder and such) process subsequent instructions. When RET instructions is being executed, few next instructions are already loaded into cpu pipeline. It's a guess, but you can start digging here and if you find out (maybe the specific number of NOPs that are safe, share your findings please.

How does a compiled "Hello World" C program store the String using machine language?

so I've started learning about machine language today. I wrote a basic "Hello World" program in C which prints "Hello, world!" ten times using a for loop. I then used the Gnu Debugger to disassemble main and look at the code in machine language (my computer has a x86 processor and I've set gdb up to use intel syntax):
user#PC:~/Path/To/Code$ gdb -q ./a.out
Reading symbols from ./a.out...done.
(gdb) list
1 #include <stdio.h>
2
3 int main()
4 {
5 int i;
6 for(i = 0; i < 10; i++) {
7 printf("Hello, world!\n");
8 }
9 return 0;
10 }
(gdb) disassemble main
Dump of assembler code for function main:
0x0804841d <+0>: push ebp
0x0804841e <+1>: mov ebp,esp
0x08048420 <+3>: and esp,0xfffffff0
0x08048423 <+6>: sub esp,0x20
0x08048426 <+9>: mov DWORD PTR [esp+0x1c],0x0
0x0804842e <+17>: jmp 0x8048441 <main+36>
0x08048430 <+19>: mov DWORD PTR [esp],0x80484e0
0x08048437 <+26>: call 0x80482f0 <puts#plt>
0x0804843c <+31>: add DWORD PTR [esp+0x1c],0x1
0x08048441 <+36>: cmp DWORD PTR [esp+0x1c],0x9
0x08048446 <+41>: jle 0x8048430 <main+19>
0x08048448 <+43>: mov eax,0x0
0x0804844d <+48>: leave
0x0804844e <+49>: ret
End of assembler dump.
(gdb) x/s 0x80484e0
0x80484e0: "Hello, world!"
I understand most of the machine code and what each of the commands do. If I understood it correctly, the address "0x80484e0" is loaded into the esp register so that can use the memory at this address. I examined the address, and to no surprise it contained the desired string. My question now is - how did that string get there in the first place? I can't find a part in the program that sets the string up at this location.
I also don't understand something else: When I first start the program, the eip points to , where the variable i is initialized at [esp+0x1c]. However, the address that esp points to is changed later on in the program (to 0x80484e0), but [esp+0x1c] is still used for "i" after that change. Shouldn't the adress [esp+0x1c] change when the address esp points to changes?
I binary or program is made up of both machine code and data. In this case your string which you put in the source code, the compiler too that data which is just bytes, and because of how it was used was considered read only data, so depending on the compiler that might land in .rodata or .text or some other name the compiler might use. Gcc would probably call it .rodata. The program itself is in .text. The linker comes along and when it links things finds a place for .text, .data, .bss, .rodata, and any other items you may have and then connects the dots. In the case of your call to printf the linker knows where it put the string, the array of bytes, and it was told what its name was (some internal temporary name no doubt) and the printf call was told about that name to so the linker patches up the instruction to grab the address to the format string before calling printf.
Disassembly of section .text:
0000000000400430 <main>:
400430: 53 push %rbx
400431: bb 0a 00 00 00 mov $0xa,%ebx
400436: 66 2e 0f 1f 84 00 00 nopw %cs:0x0(%rax,%rax,1)
40043d: 00 00 00
400440: bf e4 05 40 00 mov $0x4005e4,%edi
400445: e8 b6 ff ff ff callq 400400 <puts#plt>
40044a: 83 eb 01 sub $0x1,%ebx
40044d: 75 f1 jne 400440 <main+0x10>
40044f: 31 c0 xor %eax,%eax
400451: 5b pop %rbx
400452: c3 retq
400453: 66 2e 0f 1f 84 00 00 nopw %cs:0x0(%rax,%rax,1)
40045a: 00 00 00
40045d: 0f 1f 00 nopl (%rax)
Disassembly of section .rodata:
00000000004005e0 <_IO_stdin_used>:
4005e0: 01 00 add %eax,(%rax)
4005e2: 02 00 add (%rax),%al
4005e4: 48 rex.W
4005e5: 65 6c gs insb (%dx),%es:(%rdi)
4005e7: 6c insb (%dx),%es:(%rdi)
4005e8: 6f outsl %ds:(%rsi),(%dx)
4005e9: 2c 20 sub $0x20,%al
4005eb: 77 6f ja 40065c <__GNU_EH_FRAME_HDR+0x68>
4005ed: 72 6c jb 40065b <__GNU_EH_FRAME_HDR+0x67>
4005ef: 64 21 00 and %eax,%fs:(%rax)
the compiler will have encoded this instruction but left the address as zeros probably or some fill
400440: bf e4 05 40 00 mov $0x4005e4,%edi
so that the linker could fill it in later. The gnu disassembler attempts to disassemble the .rodata (and .data, etc) blocks which doesnt make sense, so ignore the instructions it is trying to interpret your string which starts at address 0x4005e4.
Before linking a disassembly of the object shows the two sections .text and .rodata
Disassembly of section .text.startup:
0000000000000000 <main>:
0: 53 push %rbx
1: bb 0a 00 00 00 mov $0xa,%ebx
6: 66 2e 0f 1f 84 00 00 nopw %cs:0x0(%rax,%rax,1)
d: 00 00 00
10: bf 00 00 00 00 mov $0x0,%edi
15: e8 00 00 00 00 callq 1a <main+0x1a>
1a: 83 eb 01 sub $0x1,%ebx
1d: 75 f1 jne 10 <main+0x10>
1f: 31 c0 xor %eax,%eax
21: 5b pop %rbx
22: c3 retq
0000000000000000 <.rodata.str1.1>:
0: 48 rex.W
1: 65 6c gs insb (%dx),%es:(%rdi)
3: 6c insb (%dx),%es:(%rdi)
4: 6f outsl %ds:(%rsi),(%dx)
5: 2c 20 sub $0x20,%al
7: 77 6f ja 78 <main+0x78>
9: 72 6c jb 77 <main+0x77>
b: 64 21 00 and %eax,%fs:(%rax)
unlinked it has to just pad this address/offset for the linker to fill in later.
10: bf 00 00 00 00 mov $0x0,%edi
also note the object contains only the string in .rodata. linking with libraries and other items to make it a complete program clearly added more .rodata, but the linker manages all of that.
Perhaps easier to see with this example
void more_fun ( unsigned int, unsigned int, unsigned int );
unsigned int a;
unsigned int b=5;
const unsigned int c=7;
void fun ( void )
{
more_fun(a,b,c);
}
disassembled as a object
Disassembly of section .text:
0000000000000000 <fun>:
0: 8b 35 00 00 00 00 mov 0x0(%rip),%esi # 6 <fun+0x6>
6: 8b 3d 00 00 00 00 mov 0x0(%rip),%edi # c <fun+0xc>
c: ba 07 00 00 00 mov $0x7,%edx
11: e9 00 00 00 00 jmpq 16 <fun+0x16>
Disassembly of section .data:
0000000000000000 <b>:
0: 05 .byte 0x5
1: 00 00 add %al,(%rax)
...
Disassembly of section .rodata:
0000000000000000 <c>:
0: 07 (bad)
1: 00 00 add %al,(%rax)
...
and for whatever reason you have to link it to see the .bss section. The point of the example is the machine code for the function is in .text, the uninitialized global is in .bss, the initialized global is .data and the const initialized global is .rodata. The compiler was smart enough to know that a const even if it is global wont change so it can just hardcode that value into the math and not need to read from ram, but the other two variables it has to read from ram so generates an instruction with the address zeros to be filled in by the linker at link time.
In your case your read only/const data was a collection of bytes and it wasnt a math operation so the bytes as defined in your source file were placed in memory so they could be pointed at as the first parameter to printf.
There is more to a binary than just machine code. And the compiler and linker can have things placed in memory for the machine code to get, the machine code itself does not have to write every value that will be used by the rest of the machine code.
The compiler 'hard wires' the string into the object code and the linker then 'hard wires' it into the machine code.
Not that the string is embedded into the code, and not stored in a data area meaning that if you took a pointer to the string and attempted to change it you would get an exception.

Wnh does gcc generate extra xchg instructions, and one nop at the end? [duplicate]

I've been working with C for a short while and very recently started to get into ASM. When I compile a program:
int main(void)
{
int a = 0;
a += 1;
return 0;
}
The objdump disassembly has the code, but nops after the ret:
...
08048394 <main>:
8048394: 55 push %ebp
8048395: 89 e5 mov %esp,%ebp
8048397: 83 ec 10 sub $0x10,%esp
804839a: c7 45 fc 00 00 00 00 movl $0x0,-0x4(%ebp)
80483a1: 83 45 fc 01 addl $0x1,-0x4(%ebp)
80483a5: b8 00 00 00 00 mov $0x0,%eax
80483aa: c9 leave
80483ab: c3 ret
80483ac: 90 nop
80483ad: 90 nop
80483ae: 90 nop
80483af: 90 nop
...
From what I learned nops do nothing, and since after ret wouldn't even be executed.
My question is: why bother? Couldn't ELF(linux-x86) work with a .text section(+main) of any size?
I'd appreciate any help, just trying to learn.
First of all, gcc doesn't always do this. The padding is controlled by -falign-functions, which is automatically turned on by -O2 and -O3:
-falign-functions
-falign-functions=n
Align the start of functions to the next power-of-two greater than n, skipping up to n bytes. For instance,
-falign-functions=32 aligns functions to the next 32-byte boundary, but -falign-functions=24 would align to the next 32-byte boundary only
if this can be done by skipping 23 bytes or less.
-fno-align-functions and -falign-functions=1 are equivalent and mean that functions will not be aligned.
Some assemblers only support this flag when n is a power of two; in
that case, it is rounded up.
If n is not specified or is zero, use a machine-dependent default.
Enabled at levels -O2, -O3.
There could be multiple reasons for doing this, but the main one on x86 is probably this:
Most processors fetch instructions in aligned 16-byte or 32-byte blocks. It can be
advantageous to align critical loop entries and subroutine entries by 16 in order to minimize
the number of 16-byte boundaries in the code. Alternatively, make sure that there is no 16-byte boundary in the first few instructions after a critical loop entry or subroutine entry.
(Quoted from "Optimizing subroutines in assembly
language" by Agner Fog.)
edit: Here is an example that demonstrates the padding:
// align.c
int f(void) { return 0; }
int g(void) { return 0; }
When compiled using gcc 4.4.5 with default settings, I get:
align.o: file format elf64-x86-64
Disassembly of section .text:
0000000000000000 <f>:
0: 55 push %rbp
1: 48 89 e5 mov %rsp,%rbp
4: b8 00 00 00 00 mov $0x0,%eax
9: c9 leaveq
a: c3 retq
000000000000000b <g>:
b: 55 push %rbp
c: 48 89 e5 mov %rsp,%rbp
f: b8 00 00 00 00 mov $0x0,%eax
14: c9 leaveq
15: c3 retq
Specifying -falign-functions gives:
align.o: file format elf64-x86-64
Disassembly of section .text:
0000000000000000 <f>:
0: 55 push %rbp
1: 48 89 e5 mov %rsp,%rbp
4: b8 00 00 00 00 mov $0x0,%eax
9: c9 leaveq
a: c3 retq
b: eb 03 jmp 10 <g>
d: 90 nop
e: 90 nop
f: 90 nop
0000000000000010 <g>:
10: 55 push %rbp
11: 48 89 e5 mov %rsp,%rbp
14: b8 00 00 00 00 mov $0x0,%eax
19: c9 leaveq
1a: c3 retq
This is done to align the next function by 8, 16 or 32-byte boundary.
From “Optimizing subroutines in assembly language” by A.Fog:
11.5 Alignment of code
Most microprocessors fetch code in aligned 16-byte or 32-byte blocks. If an importantsubroutine entry or jump label happens to be near the end of a 16-byte block then themicroprocessor will only get a few useful bytes of code when fetching that block of code. Itmay have to fetch the next 16 bytes too before it can decode the first instructions after thelabel. This can be avoided by aligning important subroutine entries and loop entries by 16.
[...]
Aligning a subroutine entry is as simple as putting as many
NOP
's as needed before thesubroutine entry to make the address divisible by 8, 16, 32 or 64, as desired.
As far as I remember, instructions are pipelined in cpu and different cpu blocks (loader, decoder and such) process subsequent instructions. When RET instructions is being executed, few next instructions are already loaded into cpu pipeline. It's a guess, but you can start digging here and if you find out (maybe the specific number of NOPs that are safe, share your findings please.

i386-elf-gcc out put strange assembler command about "static a = 0"

i am write a mini os. And when i write this code to show time clock, its goes wrong
7 void timer_callback(pt_regs *regs)
8 {
9 static uint32_t tick = 0;
10 printf("Tick: %dtimes\n", tick);
11 tick++;
12 }
tick is initialise not with 0, but 1818389861. but if tick init with 0x01 or anything else zero, it's ok!!!
so i wirte a simple c file then objdump:
staic.o: file format elf32-i386
Disassembly of section .text:
00000000 <main>:
extern void printf(char *, int);
int main(){
0: 8d 4c 24 04 lea 0x4(%esp),%ecx
4: 83 e4 f0 and $0xfffffff0,%esp
7: ff 71 fc pushl -0x4(%ecx)
a: 55 push %ebp
b: 89 e5 mov %esp,%ebp
d: 51 push %ecx
e: 83 ec 04 sub $0x4,%esp
static int a = 1;
printf("%d\n", a);
11: a1 00 00 00 00 mov 0x0,%eax
16: 83 ec 08 sub $0x8,%esp
19: 50 push %eax
1a: 68 00 00 00 00 push $0x0
1f: e8 fc ff ff ff call 20 <main+0x20>
24: 83 c4 10 add $0x10,%esp
return 0;
27: b8 00 00 00 00 mov $0x0,%eax
}
2c: 8b 4d fc mov -0x4(%ebp),%ecx
2f: c9 leave
30: 8d 61 fc lea -0x4(%ecx),%esp
33: c3 ret
so strange, no memory used!!!
Update: let me say it clearly
the second static.c is an experiment, it was thought it show no memory used, but i was wrong, mov 0x0 %eab is. i confuse 0x0 and $0x0 /..\
my origin problem is why tick not succeed init with 0.(but can init with 1 or anyelsenumber).
i look up it again use gdb, ok, it do use memory like mov
eax,ds:0x106010,but the real strong thing is the memory x 0x106010 is not 0,but it should be, just as i said, if i let tick = 1 or anythingelse, memory do init as i want, that is the strange thing!
the tool: gdb ,objdump return different asm(different means,not formate),because, just learn os,not good at c, so i let it go,ignore it....
Memory is used, be sure of that; however, you won't find that memory in the .text section. Memory for static variables is allocated in either .bss (when zero-initialized; or, in case of C++, dynamically initialized) or .data (when non-zero initialized) section.
When dumping object files with objdump using the -d (disassembly) option, it is important to also use the -r (relocations) option. Without that, the disassembly you get is deceiving and makes little sense.
In your case, the instruction at addresses 11 and 1f must have relocations, at address 11, to the variable a and at address 1f, to the function printf. The instruction at address 11 loads the value from your variable a, without proper relocations it looks as if it loaded a value from address 0.
As to your original question, the value you get, 1818389861, or 0x6C626D65, is quite remarkable. I would bet that somewhere in your program you have a buffer overrun involving a string containing the subsequence embl.
As a side note, I would like to call your attention to the use of correct type specifications in printf calls. The type specification %d corresponds to the type int; on all modern mainstream architectures, int and int32_t are of the same size. However, that is not guaranteed to always be so. There are special type specifications for use with explicitly-sized types, for example, for an int32_t you use "PRId32":
uint32_t x;
printf("%"PRId32, x);

GCC 4.3/4.4 vs MSC 6 on i386 optimization for size fail

I am not sure what am I doing wrong, but I've tried reading manuals about calling conventions of GCC and found nothing useful there. My current problem is GCC generates excessively LARGE code for a very simple operation, like shown below.
main.c:
#ifdef __GNUC__
// defines for GCC
typedef void (* push1)(unsigned long);
#define PUSH1(P,A0)((push1)P)((unsigned long)A0)
#else
// defines for MSC
typedef void (__stdcall * push1)(unsigned long);
#define PUSH1(P,A0)((push1)P)((unsigned long)A0)
#endif
int main() {
// pointer to nasm-linked exit syscall "function".
// will not work for win32 target, provided as an example.
PUSH1(0x08048200,0x7F);
}
Now, let's build and dump it with gcc: gcc -c main.c -Os;objdump -d main.o:
main.o: file format elf32-i386
Disassembly of section .text:
00000000 <.text>:
0: 8d 4c 24 04 lea 0x4(%esp),%ecx
4: 83 e4 f0 and $0xfffffff0,%esp
7: ff 71 fc pushl -0x4(%ecx)
a: b8 00 82 04 08 mov $0x8048200,%eax
f: 55 push %ebp
10: 89 e5 mov %esp,%ebp
12: 51 push %ecx
13: 83 ec 10 sub $0x10,%esp
16: 6a 7f push $0x7f
18: ff d0 call *%eax
1a: 8b 4d fc mov -0x4(%ebp),%ecx
1d: 83 c4 0c add $0xc,%esp
20: c9 leave
21: 8d 61 fc lea -0x4(%ecx),%esp
24: c3 ret
That's the minimum size code I am able to get... If I don't specify -O* or specify other values, it will be 0x29 + bytes long.
Now, let's build it with ms c compiler v 6 (yea, one of year 98 iirc): wine /mnt/ssd/msc/6/cl /c /TC main.c;wine /mnt/ssd/msc/6/dumpbin /disasm main.obj:
Dump of file main.obj
File Type: COFF OBJECT
_main:
00000000: 55 push ebp
00000001: 8B EC mov ebp,esp
00000003: 6A 7F push 7Fh
00000005: B8 00 82 04 08 mov eax,8048200h
0000000A: FF D0 call eax
0000000C: 5D pop ebp
0000000D: C3 ret
How do I make GCC generate the similar by size code? any hints, tips? Don't you agree resulting code should be small as that? Why does GCC append so much useless code? I thought it'd be smarter than such old stuff like msc6 when optimizing for size. What am I missing here?
main() is special here: gcc is doing some extra work to make the stack 16-byte aligned at the entry point of the program. So the size of the result aren't directly comparable... try renaming main() to f() and you'll see gcc generates drastically different code.
(The MSVC-compiled code doesn't need to care about alignment because Windows has different rules for stack alignment.)
This is the best reference I can get. I'm on Windows now and too lazy to login to my Linux to test. Here (MinGW GCC 4.5.2), the code is smaller than yours. One difference is the calling convention, stdcall of course has a few bytes advantage over cdecl (default on GCC if not specified or with -O1 and I guess with -Os, too) to clean up the stack.
Here's the way I compile and the result (source code is purely copy pasted from your post)
gcc -S test.c:
_main:
pushl %ebp #
movl %esp, %ebp #,
andl $-16, %esp #,
subl $16, %esp #,
call ___main #
movl $127, (%esp) #,
movl $134513152, %eax #, tmp59
call *%eax # tmp59
leave
ret
gcc -c -o test.o test.c && objdump -d test.o:
test.o: file format pe-i386
Disassembly of section .text:
00000000 <_main>:
0: 55 push %ebp
1: 89 e5 mov %esp,%ebp
3: 83 e4 f0 and $0xfffffff0,%esp
6: 83 ec 10 sub $0x10,%esp
9: e8 00 00 00 00 call e <_main+0xe>
e: c7 04 24 7f 00 00 00 movl $0x7f,(%esp)
15: b8 00 82 04 08 mov $0x8048200,%eax
1a: ff d0 call *%eax
1c: c9 leave
1d: c3 ret
1e: 90 nop
1f: 90 nop

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