Why does GCC put a no-op push / pop when using int foo asm ("ebx") to associate var with a register?

Question

I am trying to write code to read from a register, and this is what I still have:

unsigned int readEBX(void) {
    register unsigned int reg asm("ebx");
    return reg;
}

The function works, but it compiles into something strange:

readEBX():
    mov  eax, ebx
    push ebx
    pop  ebx
    ret

Why do this when you use push-then-pop ebx? Doesn't that mean anything? When I replace ebxwith a different register (say eaxor ecx), then it produces cleaner code (just a retand a mov eax, ecx; retrespectively).

See this example result of Godbolt .

Answer 1

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( mov)

unsigned int readEBX(void) {
  register unsigned int ret __asm__("eax");
  __asm__ volatile("mov %%ebx, %0" : "=r"(ret));
  return ret;
}

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Answer 2

:

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, , , , gcc push/pop insns /.

asm, ebx - . , , , ebx. readEBX , asm .

Answer 3

push/pop ¹

[, reg C, ebx, ebx reg asm], , promises : , ( ) ebx, asm .

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/ , ebx , . push .

, ebx, asm,

long reg; asm("mov %%ebx,%0" : "=r"(reg));

, push/pop , , push asm, .

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