CODEWITHSUNDEEP
cgccassembly
user4016243
user4016243

Reputation: 11

Is gcc doing a recursive call?

I'm compiling this code with -O3 -x c -std=c99 -fno-builtin -nostdlib -ffreestanding

unsigned char *memset(unsigned char *dest, unsigned char val, int count)
{
    unsigned char* p = dest;
    while (count--)
        *p++ = val;

    return dest;
}

#include <stdio.h>

int main()
{
    unsigned char c[20];
    memset(c, 'a', 19);
    c[19] = '\0';
    printf((const char*) c);
}

and using godbolt to examine what memset gcc is calling in the assembly output.

memset:
    test    edx, edx
    je  .L6
    sub edx, 1
    sub rsp, 8
    movzx   esi, sil
    add rdx, 1
    call    memset
    add rsp, 8
    ret
.L6:
    mov rax, rdi
    ret
main:
    sub rsp, 40
    movabs  rax, 7016996765293437281
    mov QWORD PTR [rsp], rax
    mov QWORD PTR [rsp+8], rax
    mov eax, 24929
    mov WORD PTR [rsp+16], ax
    mov rdi, rsp
    xor eax, eax
    mov BYTE PTR [rsp+18], 97
    mov BYTE PTR [rsp+19], 0
    call    printf
    add rsp, 40
    ret

With the flags I used I'm attempting to eliminate all possibility of it calling a built-in memset and judging from the colorization godbolt uses, it looks like gcc is doing a recursive call at *p++ = val;. So is it doing recursion or calling builtin memset?

Upvotes: 1

Views: 281

Answers (1)

Sparky
Sparky

Reputation: 14057

As others have indicated, the setting of the array c elements has been inlined. As a result, the memset() you implemented is not even getting called. This is a result of the use of the -03 compiler option. The compiler is being very aggressive in its optimizations. Furthermore, there is no recursion on the execution path.

However, that does not entirely answer your question. The memset() shown in the disassembled output is indeed NOT the built in version and it is not even being executed.

Incidentally, you do not need to apply the -fno-builtin flag as the -ffreestanding flag automatically implies it. Also, if you enable garbage collection, I am sure that will find that the memset() routine in the disassembled output will vanish.

Upvotes: 1

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