When I do int a = std::move(b) (b is int also), is it same as just a = b?

Question

When I do int a = std::move(b) (b is int also), is it same as just a = b?

Answer 1

Depends on the compiler! The assembler for the variant with std::move without optimization will try to remove the "reference" even though it is unnecessary, which the ASM code for the variant without std::move will not - this gives you a slight overhead (a call to std::move which contains a few instructions and an additional movl on the top level) in terms of CPU instructions!

Test Code:

Example without optimization using GCC 8.2 in X86_64 assembler:

#include <stdio.h>

int main()
{
    c = b;
    return 0;
}


int alternative()
{
    c = std::move(b);
    return 0;
}

Assembler O0:

main:
        pushq   %rbp
        movq    %rsp, %rbp
        movl    b(%rip), %eax
        movl    %eax, c(%rip)
        movl    $0, %eax
        popq    %rbp
        ret

alternative():
        pushq   %rbp
        movq    %rsp, %rbp
        movl    $b, %edi
        call    std::remove_reference<int&>::type&& std::move<int&>(int&)
        movl    (%rax), %eax
        movl    %eax, c(%rip)
        movl    $0, %eax
        popq    %rbp
        ret

std::remove_reference<int&>::type&& std::move<int&>(int&):
        pushq   %rbp
        movq    %rsp, %rbp
        movq    %rdi, -8(%rbp)
        movq    -8(%rbp), %rax
        popq    %rbp
        ret

However, if you turn on optimization (-O3), it really becomes the same in terms of CPU instructions:

main:
        movl    b(%rip), %eax
        movl    %eax, c(%rip)
        xorl    %eax, %eax
        ret

alternative():
        movl    b(%rip), %eax
        movl    %eax, c(%rip)
        xorl    %eax, %eax
        ret