Is mov %esi, %esi a no-op or not on x86-64?

mov %esi, %esi

zeros out the high 32 bits of %rsi, and is therefore not a no-op on x86_64.

See Why do x86-64 instructions on 32-bit registers zero the upper part of the full 64-bit register?

#include <stdio.h>

int main(int argc, char * argv[])
{
    void * reg_rsi = 0;

    asm (
        "movq $0x1234567812345678, %%rsi;\n"
        "movl %%esi, %%esi;\n"
        "movq %%rsi, %0;\n"
        : "=r" (reg_rsi)
        : /* no inputs */
        : /* no clobbered */
    );

    printf("reg_rsi = %p\n", reg_rsi);

    return 0;
}

This gives "reg_rsi = 0x12345678" for my x86_64 machine.