I'm trying to understand exactly how Google's DoNotOptimize() is supposed to work.
For completeness, here is its definition (for clang, and non-const data):
template <class Tp>
inline BENCHMARK_ALWAYS_INLINE void DoNotOptimize(Tp& value) {
asm volatile("" : "+r,m"(value) : : "memory");
}
As I understand we can use this in code like this:
start_time = time();
bench_output = run_bench(bench_inputs);
result = time() - start_time;
To ensure that the benchmark stays in the critical section:
start_time = time();
DoNotOptimize(bench_inputs);
bench_output = run_bench(bench_inputs);
DoNotOptimise(bench_output);
result = time() - start_time;
Specifically what I don't understand is why this guarantees (does it?) that run_bench() is not moved above start_time = time().
(Someone asked exactly this in this comment, however I don't understand the answer).
As I understand, the above DoNotOptimze() does several things:
It forces value to the stack, as it is passed by C++ reference. You can't have a pointer to a register, so it must be in memory.
Because value is now on the stack, subsequently clobbering memory (as done in the asm constraints) will force the compiler to assume that value is both read and written by the call to DoNotOptimize(value).
(it's not clear to me if the +r,m constraint is relevant. As far as I know this says that the pointer itself may be stored in a register or in memory, but the pointer value itself may be read and/or written.)
And this is where things get fuzzy for me.
If start_time is also stack allocated, the memory clobbering in DoNotOptimize() will mean that the compiler must assume that DoNotOptimize() potentially reads start_time. Therefore the order of the statements can only be:
start_time = time(); // on the stack
DoNotOptimize(bench_inputs); // reads start_time, writes bench_inputs
bench_output = run_bench(bench_inputs)
But if start_time is not stored in memory, but instead in a register, then clobbering memory will not clobber start_time, right? In that case the desired ordering of start_time = time() and DoNotOptimize(bench_inputs) is lost and the compiler is free to do:
DoNotOptimize(bench_inputs); // only writes bench_inputs
bench_output = run_bench(bench_inputs)
start_time = time(); // in a register
Obviously I've misunderstood something. Can anyone help explain? Thanks :)
I'm wondering if this is because reordering optimisations happen prior to register allocation, and thus everything is assumed to be stack allocated at that time. But if that were the case, then DoNotOptimize() would be redundant, as ClobberMemory() would be sufficient.
Summary: DoNotOptimize is ordered wrt. time() by the the "memory" clobbers, as if it were another function call to an opaque function that could modify any global state.
DoNotOptimize is ordered wrt. the computation of output from input by the data dependency of the calculation on the input, and the output on the calculation, as Chandler Carruth explained in the Q&A you linked. The "memory" clobber is irrelevant for this part.
"memory" clobber is like a non-inline function call
DoNotOptimize's asm statement contains a "memory" clobber. As far as the optimizer is concerned, that's equivalent to an opaque function call: it has to be assumed to read and write every globally-reachable object1. (Even ones this compilation unit might not know about.)
Since time() itself doesn't have an inline definition in any header, it can't reorder with DoNotOptimize at compile time for the same reason that a compiler can't reorder calls to foo() and bar() when it can't see the definitions of those functions. Same reason compilers don't need any special logic to stop them from reordering puts("hi"); puts("mom");.
(A hypothetical time() that could inline and only contained an asm statement would have to use asm volatile to make sure repeated calls didn't just use the first one's output. asm volatile statements can't reorder with each other or accesses to volatile variables, so that would be ok too, for a different reason.)
Footnote 1: Globally reachable = any object that might be pointed-to by any hypothetical global variable. i.e. anything except local variables within this function, or memory freshly allocated with new, if escape analysis can prove that nothing outside this function could have pointers to them.
How the asm statement works
I think you're pretty seriously misunderstanding how the asm works. "+r,m" tells the compiler to materialize the value in a register (or memory if it wants), and then use the value there at the end of the (empty) asm template as the new value of that C++ object.
So it forces the compiler to actually materialize (produce) the value somewhere, which means it has to be computed. And it means has to forget what it previously knew about the value (e.g. that it was a compile time constant 5, or non-negative, or anything) because the "+" modifier declares a read/write operand.
The point of DoNotOptimize on the input is to defeat constant-propagation that would let the benchmark optimize away.
And on the output to make sure a final result is actually materialized in a register (or memory) instead of optimizing away all the computation leading to an unused result. (This is where being asm volatile is relevant; defeating constant-propagation still works with non-volatile asm.)
So the computation you want to benchmark has to happen between the two DoNotOptimize() statements, and separately those two statements can't reorder with time().
The compiler has to assume that the asm statement modifies the value like val ^= random for all it knows, along with changing the value in memory of any/every other object except for private locals that weren't operands, so e.g. the "memory" clobber doesn't stop the compiler from keeping a local loop counter in memory. (It doesn't special case an empty asm template string here; programs don't contain asm statements like this by accident so nobody wants them optimized away.)
Misconceptions about the reference arg and picking "m"
I only got part way into the details of your attempt to reason about the "+r,m" operand and the reference function-arg before deciding it would probably be better to just explain from scratch. The correct reason isn't that complicated. But a couple things are worth specifically correcting:
The C++ function containing the asm statement can inline, letting the by-reference function arg optimize away. (It's even declared inline __attribute__((always_inline)) to force inlining even with optimization disabled, although in that case the reference variable won't optimize away.)
The net result is as if the asm statement were used directly on the C++ variable passed to DoNotOptimize. e.g. DoNotOptimize(foo) is like asm volatile("" : "+r,m"(foo) :: "memory")
The compiler can always pick register if it wants to, e.g. choosing to load a variable's value into a register before an asm statement. (And if the C++ semantics demand updating the variable's value in memory, also emitting a store instruction after the asm statement.)
For example, we can see that GCC does choose to do that. (I guess I could have used incl %0 as the example, but I just chose nop as a way to show what the compiler picked for the operand location as an alternative to # %0 pure comment, so the Godbolt compiler explorer wouldn't filter it out.)
void foo(int *p)
{
asm volatile("nop # operand picked %0" : "+r,m" (p[4]) );
}
# GCC 11.2 -O2
foo(int*):
movl 16(%rdi), %eax
nop # operand picked %eax
movl %eax, 16(%rdi)
ret
vs. clang choosing to leave the value in memory, so every instruction in the asm template would be accessing memory instead of a register. (If there were any instructions).
# clang 12.0.1 -O2 -fPIE
foo(int*): # #foo(int*)
nop # operand picked 16(%rdi)
retq
Fun fact: "r,m" is an attempt to work around a clang missed-optimization bug that makes it always pick memory for "rm" constraints, even if the value was already in a register. Spilling it first, even if it has to invent a temporary location for the value of an expression as an input.
i will write my assumptions (based on my researches) in the question below i assume that there are mistakes in my assemptions outside the question it self:
i'm looking into some code written for ARM:
this function (taken from FreeRTOS port code):
portFORCE_INLINE static uint32_t ulPortRaiseBASEPRI(void)
{
uint32_t ulOriginalBASEPRI, ulNewBASEPRI;
__asm volatile(" mrs %0, basepri \n"
" mov %1, %2 \n"
" msr basepri, %1 \n"
" isb \n"
" dsb \n"
: "=r"(ulOriginalBASEPRI), "=r"(ulNewBASEPRI)
: "i"(configMAX_SYSCALL_INTERRUPT_PRIORITY));
/* This return will not be reached but is necessary to prevent compiler
warnings. */
return ulOriginalBASEPRI;
}
i understand in gcc "=r" is output operand. so we save values from asm to C variable
now the code in my understanding is equivalent to:
ulOriginalBASEPRI = basepri
ulNewBASEPRI = configMAX_SYSCALL_INTERRUPT_PRIORITY
basepri = ulNewBASEPRI
i understand we are returning the original value of BASEPRI so thats the first line. however, i didn't understand why we assign variable ulNewBASEPRI then we use it in MSR instruction..
so I've looked in the ARMV7 instruction set and i saw this:
i assume there is no (MSR immediate) in thumb instruction and "Encoding A1" means its only in Arm instruction mode.
so we have to use =r output operand to let asembler to auto select a register for our variable am i correct?
EDIT: ignore this section because i miscounted colons
: "i"(configMAX_SYSCALL_INTERRUPT_PRIORITY));
from my understanding for assembly template:
asm ( assembler template
: output operands /* optional */
: input operands /* optional */
: list of clobbered registers /* optional */
);
isn't "i" just means (immediate) or constant in the assembly?
does this mean the third colon is not only for clobber list?
if that so, isn't it more appropriate to find the constraint "i" in the input operands?
EDIT: ignore this section because i miscounted colons
i understand isb, dsb are memory barrier stuff but i really dont understand the discription of them. what they really do?
what happen if we remove dsb or isb instruction for example.?
so we have to use =r output operand to let assembler to auto select a register for our variable am i correct?
Yes, but it's the compiler that does register allocation. It just fills in the %[operand] in the asm template string as a text substitution and feeds that to the assembler.
Alternatively, you could hard-code a specific register in the asm template string, and use a register-asm local variable to make sure an "=r" constraint picked it. Or use an "=m" memory output operand and str a result into it, and declare a clobber on any registers you used. But those alternatives are obviously terrible compared to just telling the compiler about how your block of asm can produce an output.
I don't understand why the comment says the return statement doesn't run:
/* This return will not be reached but is necessary to prevent compiler
warnings. */
return ulOriginalBASEPRI;
Raising the basepri (ARM docs) to a higher number might allow an interrupt handler to run right away, before later instructions, but if that exception ever returns, execution will eventually reach the C outside the asm statement. That's the whole point of saving the old basepri into a register and having an output operand for it, I assume.
(I had been assuming that "raise" meant higher number = more interrupts allowed. But Ross comments that it will never allow more interrupts; they're "raising the bar" = lower number = fewer interrupts allowed.)
If execution really never comes out the end of your asm, you should tell the compiler about it. There is asm goto, but that needs a list of possible branch targets. The GCC manual says:
GCC assumes that asm execution falls through to the next statement (if this is not the case, consider using the __builtin_unreachable() intrinsic after the asm statement).
Failing to do this might lead to the compiler planning to do something after the asm, and then it never happening even though in the source it's before the asm.
It might be a good idea to use a "memory" clobber to make sure the compiler has memory contents in sync with the C abstract machine. (At least for variables other than locals, which an interrupt handler might access). This is usually desirable around asm barrier instructions like dsb, but it seems here we maybe don't care about being an SMP memory barrier, just about consistent execution after changing basepri? I don't understand why that's necessary, but if you do then worth considering one way or another whether compile-time reordering of memory access around the asm statement is or isn't a problem.
You'd use a third colon-separated section in the asm statement (after the inputs) : "memory"
Without that, compilers might decide to do an assignment after this asm instead of before, leaving a value just in registers.
// actual C source
global_var = 1;
uint32_t oldpri = ulPortRaiseBASEPRI();
global_var = 2;
could optimize (via dead-store elimination) into asm that worked like this
// possible asm
global_var = 2;
uint32_t oldpri = ulPortRaiseBASEPRI();
// or global_var = 2; here *instead* of before the asm
Concerning ARM/Thumb instruction set differences on msr: you should be able to answer this yourself from the documentation. ;-) It is just 2 pages later. Edit: Chapter A8.1.3 of the linked manual clearly states how encodings are documented on instructions.
dsb (data synchronization barrier) makes sure that all memory accesses are finished before the next instruction is executed. This is really shortly written, for the full details you need to read the documentation. If you have further specific questions about this operation, please post another question.
isb (instruction synchronization barrier) purges the instruction pipeline. This pipeline buffers instructions which are already fetched from memory but are not yet executed. So the next instruction will be fetched with possibly changed memory access, and this is what a programmer expects. The note above applies here, too.
Recently I came across Assembly language. x86 assembly has an xchg instruction which swaps the contents of two registers.
Since every C code is first converted to Assembly, it would have been nice if there was a swap function inbuilt in C like in the header stdio.h. Then whenever the compiler detects the swap function, it could add the xchg directive in the assembly file.
So why this swap function was not implemented in C?
C is a cross-platform language. Assembly is architecture specific. Not every architecture has such an instruction. Moreover, C, as a high-level language doesn't have to correspond to the machine-level instruction set and features, as it's purpose is to bridge between the "human" language and the machine language, not to mimic it. Said that, a C compiler for this specific architecture might have an extension for this swapping instruction or optimize the swapping code to use this instruction if smart enough.
There are two points which can explain why swap() is not in C
1. Function call semantics:
Including a swap() function would break a very fundamental design decision in C: swap() can only work with pass-by-reference semantics (which C++ added to the language, but which are absent in C), not with pass-by-value.
2. Diversity of available assembler instructions
Apart from that, there is usually quite a number of assembler instructions on any given CPU architecture which are totally inaccessible from pure C. This includes instructions as diverse as interrupt handling instructions, virtual memory space manipulating instructions, I/O instructions, bit fiddling instructions (google the PPC instruction rlwimi for an especially powerful example of this), etc.
It is simply impossible to include any significant number of these in a general purpose language like C.
Some of these are crucial for implementing operating systems, which is why any OS must include at the very least some small amounts of assembler code. They are usually encapsulated in some functions with inline assembler or defined in the kernel headers as preprocessor directives. Other instructions are less important, or only good for optimizations, these may be generated by optimizing compilers, and many compilers do generate them (the whole class of vector functions fall in this category).
In the face of this vast diversity, the designers of C just had to cut it somewhere. And they opted for providing whatever is representable as simple operators like (+, -, ~, &, |, !, &&, ||, etc.), but did not provide anything that would require function call syntax like the swap() function you propose.
That would work for variables that fit in the register and are in the register. It would not work for large struct or variables held in memory (If you load a variable A in reg X and another, say B in reg Y, and swap them, you could skip the swapping and load A in Y and B in X directly).
Having said said, nothing prevent the compiler for a given architecture to use the swap instruction to compile:
int a;
int b;
int tmp;
tmp=a;
a=b;
b=tmp;
... If those happens to be in registers: the fact that it is not in C does not mean the compiler does not use it.
Besides what the other correct answers say, another part of your premise is wrong.
Only a really dumb compiler would want to actually emit xchg every time the source swapped variables, whether there's an intrinsic or operator for it or not. Optimizing compilers don't just transliterate C into asm, they typically convert to an SSA internal representation of the program logic, and optimize that so they can implement it with as few instructions as possible (or really in the most efficient way possible; using multiple fast instructions can be better than a single slower one).
xchg is rarely faster than 3 mov instructions, and a good compiler can simply change its local-variable <-> CPU register mapping without emitting any asm instructions in many cases. (Or inside a loop, unrolling can often optimize away swapping.) Often you need only 1 or mov instructions in asm, not all 3. e.g. if only one of the C vars being swapped needs to stay in the same register, you can do:
# start: x in EAX, y in ECX
mov edx, eax
mov eax, ecx
# end: y in EAX, x in EDX
See also Why is XCHG reg, reg a 3 micro-op instruction on modern Intel architectures?
Also note that xchg [mem], reg is atomic (implicit lock prefix), and thus is a full memory barrier, and much slower than 3 mov instructions, and with much higher impact on surrounding code because of the memory-barrier effect.
If you do actually need to exchange registers, 3x mov is pretty good. Often better than xchg reg,reg because of mov elimination, at the cost of more code-size and a tmp reg.
There's a reason compilers never use xchg. If xchg was a win, compilers would look for it as a peephole optimization the same way they look for inc eax over add eax,1, or xor eax,eax instead of mov eax,0. But they don't.
(semi-related: swapping 2 registers in 8086 assembly language(16 bits))
Even though xchg is a very elementary instruction, this doesn't mean C must have its equivalent. The fact that C sometimes maps directly to assembly is not very relevant; the standard says nothing about "assembly" (why map to assembly and not another low-level language?).
You might also ask: Why does C not have built-in vector instructions? They're becoming largely available!
There's also compiler's help: swapping variables is a very visible pattern, so such optimization shouldn't be hard to implement. And you also have inline asm, should you need it.
I'm optimizing C code for OpenRISC and I want to manually prereserve some computed values in registers, the pseudocode looks like that:
external loop
compute eight values (heavy calculations)
internal loop
use values computed above
When I looked at GCC ABI for OpenRISC I saw two groups of registers: callee-saved and temporary? Which registers I should use to store these eight values? I mean, which registers I can put on clobbered list in inline asm?
I need to hardoce registers, because we run executables on custom OpenRISC.
The answer is: whatever you like.
If you use callee-save registers then the compiler will save them for you (as long as you do mark them as clobbered).
If you use temporary registers (a.k.a. caller-save) then the compiler will be forced to save them if you make a function call. Beware that the compiler also prefers to use these for other variables, so if you use up the caller-save ones it'll have to use callee-save for other things, so it might end up being much the same difference.
At the end of the day, if you are doing heavy calculations then saving a few registers to stack before you start is not going to be a big deal.
There are some registers that contain important values (such as stack pointer) that you must not overwrite. Others, such as the GOT table pointer are less important, and the compiler will restore the value when you're done (just be sure you don't need it during the process.
Really though, you don't need to work it out for yourself: the compiler can select registers for you:
int a, b, c;
asm volatile ("whatever" : "=&w" (a), "=&w" (b), "=&w" (c));
The variables are not needed, but they must have registers assigned, so they effectively reserve a register for whatever you want. The & indicates an "early-clobber", which means that they can't share the same register as an input register (not that my example shows any).
What does the register keyword do in C language? I have read that it is used for optimizing but is not clearly defined in any standard. Is it still relevant and if so, when would you use it?
It's a hint to the compiler that the variable will be heavily used and that you recommend it be kept in a processor register if possible.
Most modern compilers do that automatically, and are better at picking them than us humans.
I'm surprised that nobody mentioned that you cannot take an address of register variable, even if compiler decides to keep variable in memory rather than in register.
So using register you win nothing (anyway compiler will decide for itself where to put the variable) and lose the & operator - no reason to use it.
It tells the compiler to try to use a CPU register, instead of RAM, to store the variable. Registers are in the CPU and much faster to access than RAM. But it's only a suggestion to the compiler, and it may not follow through.
I know this question is about C, but the same question for C++ was closed as a exact duplicate of this question. This answer therefore may not apply for C.
The latest draft of the C++11 standard, N3485, says this in 7.1.1/3:
A register specifier is a hint to the implementation that the variable so declared will be heavily used. [ note: The hint can be ignored and in most implementations it will be ignored if the address of the variable is taken. This use is deprecated ... —end note ]
In C++ (but not in C), the standard does not state that you can't take the address of a variable declared register; however, because a variable stored in a CPU register throughout its lifetime does not have a memory location associated with it, attempting to take its address would be invalid, and the compiler will ignore the register keyword to allow taking the address.
I have read that it is used for optimizing but is not clearly defined in any standard.
In fact it is clearly defined by the C standard. Quoting the N1570 draft section 6.7.1 paragraph 6 (other versions have the same wording):
A declaration of an identifier for an object with storage-class
specifier register suggests that access to the object be as fast
as possible. The extent to which such suggestions are effective is
implementation-defined.
The unary & operator may not be applied to an object defined with register, and register may not be used in an external declaration.
There are a few other (fairly obscure) rules that are specific to register-qualified objects:
Defining an array object with register has undefined behavior.
Correction: It's legal to define an array object with register, but you can't do anything useful with such an object (indexing into an array requires taking the address of its initial element).
The _Alignas specifier (new in C11) may not be applied to such an object.
If the parameter name passed to the va_start macro is register-qualified, the behavior is undefined.
There may be a few others; download a draft of the standard and search for "register" if you're interested.
As the name implies, the original meaning of register was to require an object to be stored in a CPU register. But with improvements in optimizing compilers, this has become less useful. Modern versions of the C standard don't refer to CPU registers, because they no longer (need to) assume that there is such a thing (there are architectures that don't use registers). The common wisdom is that applying register to an object declaration is more likely to worsen the generated code, because it interferes with the compiler's own register allocation. There might still be a few cases where it's useful (say, if you really do know how often a variable will be accessed, and your knowledge is better than what a modern optimizing compiler can figure out).
The main tangible effect of register is that it prevents any attempt to take an object's address. This isn't particularly useful as an optimization hint, since it can be applied only to local variables, and an optimizing compiler can see for itself that such an object's address isn't taken.
It hasn't been relevant for at least 15 years as optimizers make better decisions about this than you can. Even when it was relevant, it made a lot more sense on a CPU architecture with a lot of registers, like SPARC or M68000 than it did on Intel with its paucity of registers, most of which are reserved by the compiler for its own purposes.
Actually, register tells the compiler that the variable does not alias with
anything else in the program (not even char's).
That can be exploited by modern compilers in a variety of situations, and can help the compiler quite a bit in complex code - in simple code the compilers can figure this out on their own.
Otherwise, it serves no purpose and is not used for register allocation. It does not usually incur performance degradation to specify it, as long as your compiler is modern enough.
Storytime!
C, as a language, is an abstraction of a computer. It allows you to do things, in terms of what a computer does, that is manipulate memory, do math, print things, etc.
But C is only an abstraction. And ultimately, what it's extracting from you is Assembly language. Assembly is the language that a CPU reads, and if you use it, you do things in terms of the CPU. What does a CPU do? Basically, it reads from memory, does math, and writes to memory. The CPU doesn't just do math on numbers in memory. First, you have to move a number from memory to memory inside the CPU called a register. Once you're done doing whatever you need to do to this number, you can move it back to normal system memory. Why use system memory at all? Registers are limited in number. You only get about a hundred bytes in modern processors, and older popular processors were even more fantastically limited (The 6502 had 3 8-bit registers for your free use). So, your average math operation looks like:
load first number from memory
load second number from memory
add the two
store answer into memory
A lot of that is... not math. Those load and store operations can take up to half your processing time. C, being an abstraction of computers, freed the programmer the worry of using and juggling registers, and since the number and type vary between computers, C places the responsibility of register allocation solely on the compiler. With one exception.
When you declare a variable register, you are telling the compiler "Yo, I intend for this variable to be used a lot and/or be short lived. If I were you, I'd try to keep it in a register." When the C standard says compilers don't have to actually do anything, that's because the C standard doesn't know what computer you're compiling for, and it might be like the 6502 above, where all 3 registers are needed just to operate, and there's no spare register to keep your number. However, when it says you can't take the address, that's because registers don't have addresses. They're the processor's hands. Since the compiler doesn't have to give you an address, and since it can't have an address at all ever, several optimizations are now open to the compiler. It could, say, keep the number in a register always. It doesn't have to worry about where it's stored in computer memory (beyond needing to get it back again). It could even pun it into another variable, give it to another processor, give it a changing location, etc.
tl;dr: Short-lived variables that do lots of math. Don't declare too many at once.
You are messing with the compiler's sophisticated graph-coloring algorithm. This is used for register allocation. Well, mostly. It acts as a hint to the compiler -- that's true. But not ignored in its entirety since you are not allowed to take the address of a register variable (remember the compiler, now on your mercy, will try to act differently). Which in a way is telling you not to use it.
The keyword was used long, long back. When there were only so few registers that could count them all using your index finger.
But, as I said, deprecated doesn't mean you cannot use it.
Just a little demo (without any real-world purpose) for comparison: when removing the register keywords before each variable, this piece of code takes 3.41 seconds on my i7 (GCC), with register the same code completes in 0.7 seconds.
#include <stdio.h>
int main(int argc, char** argv) {
register int numIterations = 20000;
register int i=0;
unsigned long val=0;
for (i; i<numIterations+1; i++)
{
register int j=0;
for (j;j<i;j++)
{
val=j+i;
}
}
printf("%d", val);
return 0;
}
I have tested the register keyword under QNX 6.5.0 using the following code:
#include <stdlib.h>
#include <stdio.h>
#include <inttypes.h>
#include <sys/neutrino.h>
#include <sys/syspage.h>
int main(int argc, char *argv[]) {
uint64_t cps, cycle1, cycle2, ncycles;
double sec;
register int a=0, b = 1, c = 3, i;
cycle1 = ClockCycles();
for(i = 0; i < 100000000; i++)
a = ((a + b + c) * c) / 2;
cycle2 = ClockCycles();
ncycles = cycle2 - cycle1;
printf("%lld cycles elapsed\n", ncycles);
cps = SYSPAGE_ENTRY(qtime) -> cycles_per_sec;
printf("This system has %lld cycles per second\n", cps);
sec = (double)ncycles/cps;
printf("The cycles in seconds is %f\n", sec);
return EXIT_SUCCESS;
}
I got the following results:
-> 807679611 cycles elapsed
-> This system has 3300830000 cycles per second
-> The cycles in seconds is ~0.244600
And now without register int:
int a=0, b = 1, c = 3, i;
I got:
-> 1421694077 cycles elapsed
-> This system has 3300830000 cycles per second
-> The cycles in seconds is ~0.430700
During the seventies, at the very beginning of the C language, the register keyword has been introduced in order to allow the programmer to give hints to the compiler, telling it that the variable would be used very often, and that it should be wise to keep it’s value in one of the processor’s internal register.
Nowadays, optimizers are much more efficient than programmers to determine variables that are more likely to be kept into registers, and the optimizer does not always take the programmer’s hint into account.
So many people wrongly recommend not to use the register keyword.
Let’s see why!
The register keyword has an associated side effect: you can not reference (get the address of) a register type variable.
People advising others not to use registers takes wrongly this as an additional argument.
However, the simple fact of knowing that you can not take the address of a register variable, allows the compiler (and its optimizer) to know that the value of this variable can not be modified indirectly through a pointer.
When at a certain point of the instruction stream, a register variable has its value assigned in a processor’s register, and the register has not been used since to get the value of another variable, the compiler knows that it does not need to re-load the value of the variable in that register.
This allows to avoid expensive useless memory access.
Do your own tests and you will get significant performance improvements in your most inner loops.
c_register_side_effect_performance_boost
Register would notify the compiler that the coder believed this variable would be written/read enough to justify its storage in one of the few registers available for variable use. Reading/writing from registers is usually faster and can require a smaller op-code set.
Nowadays, this isn't very useful, as most compilers' optimizers are better than you at determining whether a register should be used for that variable, and for how long.
gcc 9.3 asm output, without using optimisation flags (everything in this answer refers to standard compilation without optimisation flags):
#include <stdio.h>
int main(void) {
int i = 3;
i++;
printf("%d", i);
return 0;
}
.LC0:
.string "%d"
main:
push rbp
mov rbp, rsp
sub rsp, 16
mov DWORD PTR [rbp-4], 3
add DWORD PTR [rbp-4], 1
mov eax, DWORD PTR [rbp-4]
mov esi, eax
mov edi, OFFSET FLAT:.LC0
mov eax, 0
call printf
mov eax, 0
leave
ret
#include <stdio.h>
int main(void) {
register int i = 3;
i++;
printf("%d", i);
return 0;
}
.LC0:
.string "%d"
main:
push rbp
mov rbp, rsp
push rbx
sub rsp, 8
mov ebx, 3
add ebx, 1
mov esi, ebx
mov edi, OFFSET FLAT:.LC0
mov eax, 0
call printf
add rsp, 8
pop rbx
pop rbp
ret
This forces ebx to be used for the calculation, meaning it needs to be pushed to the stack and restored at the end of the function because it is callee saved. register produces more lines of code and 1 memory write and 1 memory read (although realistically, this could have been optimised to 0 R/Ws if the calculation had been done in esi, which is what happens using C++'s const register). Not using register causes 2 writes and 1 read (although store to load forwarding will occur on the read). This is because the value has to be present and updated directly on the stack so the correct value can be read by address (pointer). register doesn't have this requirement and cannot be pointed to. const and register are basically the opposite of volatile and using volatile will override the const optimisations at file and block scope and the register optimisations at block-scope. const register and register will produce identical outputs because const does nothing on C at block-scope, so only the register optimisations apply.
On clang, register is ignored but const optimisations still occur.
On supported C compilers it tries to optimize the code so that variable's value is held in an actual processor register.
Microsoft's Visual C++ compiler ignores the register keyword when global register-allocation optimization (the /Oe compiler flag) is enabled.
See register Keyword on MSDN.
Register keyword tells compiler to store the particular variable in CPU registers so that it could be accessible fast. From a programmer's point of view register keyword is used for the variables which are heavily used in a program, so that compiler can speedup the code. Although it depends on the compiler whether to keep the variable in CPU registers or main memory.
Register indicates to compiler to optimize this code by storing that particular variable in registers then in memory. it is a request to compiler, compiler may or may not consider this request.
You can use this facility in case where some of your variable are being accessed very frequently.
For ex: A looping.
One more thing is that if you declare a variable as register then you can't get its address as it is not stored in memory. it gets its allocation in CPU register.
The register keyword is a request to the compiler that the specified variable is to be stored in a register of the processor instead of memory as a way to gain speed, mostly because it will be heavily used. The compiler may ignore the request.