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C performance: macro vs. constant

Time:11-25

Is there any performance difference between these two different programs?

#define K 50

void main() {
    int k = K;
}
void main() {
    int k = 50;
}

CodePudding user response:

Macros in C perform token substitution before the actual compilation stage.

This means that after preprocessing, the second piece of code is exactly the same as the first.

CodePudding user response:

In this case: No, it makes 0 difference (not even at -O0!).

This is the assembler gcc 10 on ubuntu 20.04 produced on my machine:

.file   "compare_assign.c"
.text
.globl  main
.type   main, @function

main:
.LFB0:
    .cfi_startproc
    endbr64
    pushq   %rbp
    .cfi_def_cfa_offset 16
    .cfi_offset 6, -16
    movq    %rsp, %rbp
    .cfi_def_cfa_register 6
    movl    $50, -4(%rbp)
    nop
    popq    %rbp
    .cfi_def_cfa 7, 8
    ret
    .cfi_endproc
.LFE0:
    .size   main, .-main
    .ident  "GCC: (Ubuntu 10.3.0-1ubuntu1~20.04) 10.3.0"
    .section    .note.GNU-stack,"",@progbits
    .section    .note.gnu.property,"a"
    .align 8
    .long    1f - 0f
    .long    4f - 1f
    .long    5
0:
    .string  "GNU"
1:
    .align 8
    .long    0xc0000002
    .long    3f - 2f
2:
    .long    0x3
3:
    .align 8
4:

Only difference is the title of the source file (duh):

diff compare_assign_noopt.asm compare_define_noopt.asm 
1c1
<   .file   "compare_assign.c"
---
>   .file   "compare_define.c"
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