我如何告诉GCC asm，一个输入寄存器被阻塞了?

I'm trying to do a 64=32x32 multiply via the x86 mul instruction, but I only need the high dword of the result (the edx register). So naturally, I tried listing edx as an output register and eax as a clobbered register.

我正在尝试通过x86 mul指令进行64=32x32的乘法，但是我只需要结果的高dword (edx寄存器)。因此，自然地，我尝试将edx列为输出寄存器，将eax列为阻塞寄存器。

This seems natural to me, but eax is also an input register. When I try to tell GCC that eax is clobbered, it gives an error message.

这在我看来很自然，但eax也是一个输入寄存器。当我试图告诉GCC eax崩溃时，它会给出一个错误消息。

__asm__("mull\t%2" : "=d"(div10) : "%a"(UINT32_C(0x1999999A)), "r"(number)
    : "cc", "rax");

If I try that, it throws this error message:

如果我这样做，它会抛出错误信息:

divmod10.cpp:76:91: error: can’t find a register in class ‘AREG’ while reloading
‘asm’
divmod10.cpp:76:91: error: ‘asm’ operand has impossible constraints

Omitting it compiles, but breaks the code. GCC ends up relying upon eax not being clobbered, which is incorrect:

省略它会编译，但会破坏代码。GCC最终依赖于eax不被破坏，这是不正确的:

        movl    $429496730, %eax
#APP
# 76 "divmod10.cpp" 1
        mull    %esi
# 0 "" 2
#NO_APP
        movl    %edx, %esi
#APP
# 78 "divmod10.cpp" 1
        mull    %edx
# 0 "" 2
#NO_APP

How do I do what I want?

我怎么做我想做的?

1 个解决方案

#1

Just make a useless temp for the output to go into and the compiler will optimize it out. For example:

只需要对输出做一个无用的临时变量，编译器就会优化它。例如:

__asm__("mull\t%2" : "=d"(div10), "=a"((int){0})
    : "a"(UINT32_C(0x1999999A)), "r"(number) : "cc");

That's the easiest way I know to handle clobbered inputs.

这是我所知道的处理失败输入的最简单的方法。

#1