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gccassemblyx86inline-assembly

How does gcc know the register size to use in inline assembly?

I have the inline assembly code:

#define read_msr(index, buf)    asm volatile ("rdmsr" : "=d"(buf[1]), "=a"(buf[0]) : "c"(index))

The code using this macro:

u32 buf[2];

read_msr(0x173, buf);

I found the disassembly is (using gnu toolchain):

mov    eax,0x173
mov    ecx,eax
rdmsr  
mov    DWORD PTR [rbp-0xc],edx
mov    DWORD PTR [rbp-0x10],eax

The question is that 0x173 is less than 0xffff, why gcc does not use mov cx, 0x173? Will the gcc analysis the following instruction rdmsr? Will the gcc always know the correct register size?

Solution

  • It depends on the size of the value or variable passed.

    If you pass a "short int" it will set "cx" and read the data from "ax" and "dx" (if buf is a short int, too).

    For char it would access "cl" and so on.

    So "c" refers to the "ecx" register, but this is accessed with "ecx", "cx", or "cl" depending on the size of the access, which I think makes sense.

    To test you can try passing (unsigned short)0x173, it should change the code.

    There is no analysis of the inline assembly (in fact it is after text substitution direclty copied to the output assembly, including syntax errors). Also there is no default register size, depending on whether you have a 32 or 64 bit target. This would be way to limiting.