__asm __volatile 如何执行？

Question

gcc_inline void
lcr0(uint32_t val)
{
    __asm __volatile("movl %0,%%cr0" : : "r" (val));
}

In the above code, I'm not sure where val is inserted into the assembly string. Does val replace the %c in the string?

If possible can someone clarify what : : "r" does as well?

Answer 1

"r" means you are specifying %0 to be a register (as an input). (val) means you are specifying that the register should contain the value of val . So the compiler will allocate a register and make sure it contains val . For x86_64 the first argument for the function will be in %edi / %rdi , and that is what %0 will expand to.

---

I stand corrected...

...if the function is not inlined , val will be passed in edi / rdi but might be shuffled around before the asm , but the "r" will cause the compiler to arrange for it to be in some register for the asm . (See the effect of -O0, below).

Also a function which is not declared/defined to be inline may be inlined, at higher levels of optimization.

I note that it is only possible to read/write CR0 to/from a general purpose register and then only at privilege level 0. @PeterCordes notes that a "memory" clobber would probably a Good Idea. Clearly, changing CR0 can have really exciting side effects !

---

When I tried this at -O0 I found that a simple inline was ignored, and the function compiled for x86_64 to:

   lcr0:
        pushq   %rbp
        movq    %rsp, %rbp
        movl    %edi, -4(%rbp)
        movl    -4(%rbp), %eax
        movl %eax,%cr0
        nop
        popq    %rbp
        ret

I guess that gcc_inline may include __attribute__((__always_inline__)) , in which case even at -O0 lcr0 is inlined -- but with lots of lovely stack business. This time for x86:

  main:
        pushl   %ebp
        movl    %esp, %ebp
        subl    $16, %esp
        movl    12(%ebp), %eax
        movl    (%eax), %eax
        movl    (%eax), %eax
        movl    %eax, -4(%ebp)
        movl    -4(%ebp), %eax
        movl    %eax, -8(%ebp)
        movl    -8(%ebp), %eax
        movl %eax,%cr0
        nop
        movl    $0, %eax
        leave
        ret