Passing arguments to functions with const parameters: is it faster?

becko picture becko · Sep 6, 2012 · Viewed 26.6k times · Source

Consider, for example:

int sum(int a, int b)
{
    return a + b;
}

vs.

int sum(const int a, const int b)
{
    return a + b;
}

Is the second approach in general faster?

Function parameters in C are copied and sent to the function, so that changes inside the function do not affect the original values. My reasoning is that in the second sum above, the compiler knows for sure that a and b are not modified inside the function, so it can just pass the original values without copying them first. That's why I think the second sum is faster than the first. But I don't really know. In the particular simple example of sum above, the differences, if any, should be minimal.

Edit: The sum example is just to illustrate my point. I don't expect that in this particular example there should be large differences. But I wonder if in more complicated situations the const modifier inside a function parameter can be exploited by the compiler to make the function faster. I doubt that the compiler can always determine whether a parameter is changed inside a function (hence my 2nd question below); hence I'd expect that when it finds a const modifier, it does something different than when there's no const modifier.

Question: In general, a function will be faster when its arguments are const, than when they are not?

Question 2: In general, can a C compiler (theoretically) always determine whether a function parameter is changed inside the function?

Answer

Richard J. Ross III picture Richard J. Ross III · Sep 6, 2012

Short answer: No

Long answer, no, with proof.

I ran this test, a couple of times, and saw no real time difference, on my MacBook pro compiled with clang:

int add(int a, int b)
{
    return a + b;
}

const int cadd(const int a, const int b)
{
    return a + b;
}

int main (int argc, char * argv[])
{
#define ITERS 1000000000

    clock_t start = clock();
    int j = 0;
    for (int i = 0; i < ITERS; i++)
    {
        j += add(i, i + 1);
    }

    printf("add took %li ticks\n", clock() - start);

    start = clock();
    j = 0;
    for (int i = 0; i < ITERS; i++)
    {
        j += cadd(i, i + 1);
    }

    printf("cadd took %li ticks\n", clock() - start);

    return 0;
}

Output

add took 4875711 ticks
cadd took 4885519 ticks

These times really should be taken with a grain of salt, however, as clock isn't the most accurate of timing functions, and can be influenced by other running programs.

So, here is the compared assembly generated:

_add:
    .cfi_startproc
    pushq   %rbp
    .cfi_def_cfa_offset 16
    .cfi_offset %rbp, -16
    movq    %rsp, %rbp
    .cfi_def_cfa_register %rbp
    movl    %edi, -4(%rbp)
    movl    %esi, -8(%rbp)
    movl    -4(%rbp), %esi
    addl    -8(%rbp), %esi
    movl    %esi, %eax
    popq    %rbp
    ret

_cadd:                                 
    .cfi_startproc    
    pushq   %rbp
    .cfi_def_cfa_offset 16
    .cfi_offset %rbp, -16
    movq    %rsp, %rbp
    .cfi_def_cfa_register %rbp
    movl    %edi, -4(%rbp)
    movl    %esi, -8(%rbp)
    movl    -4(%rbp), %esi
    addl    -8(%rbp), %esi
    movl    %esi, %eax
    popq    %rb

So, as you can see, there is No difference between the two. Passing an argument as const is only a hint to the caller the the argument will not be changed, and in a simple scenario like the one described above, will not result in any different assembly compiled.