Assembler program to swap contents of registers

Question

I'm trying to do a real simple assembler program to swap the contents of registers. This is what i've tried:

movq (%rcx), %rax
movq (%rbx), %rdx
movq %rdx, (%rcx)
movq %rax, (%rbx)
ret

It gives me segmentation fault.

Here is an example of a working program in c:

void swap(int64_t *a, int64_t *b) {
    int64_t c = *a;
    *a = *b;
    *b = c;
}

Answer 1

See: https://en.wikipedia.org/wiki/X86_calling_conventions

You neglected to mention whether you're compiling for Microsoft/Win x64 or for the System V AMD64 ABI [or for something else entirely].

You are using AT&T asm syntax, so I'm assuming you want the SysV calling convention. (Since tools like GCC and GAS are more common on Linux / MacOS. But if you're using MinGW w64 on Windows then you'll want the Windows convention.)

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You're assuming the args are in: %rcx and %rbx. This does not correspond with either convention [although it is somewhat closer to the MS ABI]

For System V AMD64 ABI (e.g. Linux, BSD, MacOS), the first two args are passed in %rdi and %rsi respectively. And, not in %rdx and %rcx (which are for the 3rd and 4th args).

You can always use %rax and %rdx as temp regs because %rax holds the function return value and %rdx is an arg reg so caller won't expect them to be preserved.

So, you want:

# Non-Windows
movq (%rdi),%rax
movq (%rsi),%rdx
movq %rdx,(%rdi)
movq %rax,(%rsi)
ret

For MS 64 bit, the arg registers are: %rcx, %rdx, %r8, %r9

So, you'd want:

# Windows
movq (%rcx),%rax
movq (%rdx),%r8
movq %r8,(%rcx)
movq %rax,(%rdx)
ret