问题描述
我试图做一个64 = 32x32乘法通过86 MUL
指示,但我只需要结果的高DWORD(即 EDX
报名)。所以很自然,我想上市的 EDX
作为输出寄存器和 EAX
作为修饰寄存器。
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.
这似乎很自然的我,但 EAX
是也的输入寄存器。当我试着告诉GCC的 EAX
被破坏的,它提供了一个错误信息。
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.
__asm__("mull\t%2" : "=d"(div10) : "%a"(UINT32_C(0x1999999A)), "r"(number)
: "cc", "rax");
如果我尝试,它抛出此错误消息:
If I try that, it throws this error message:
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
忽略它编译,但打破了code。 GCC最终依赖于 EAX
未被破坏的,这是不正确的:
Omitting it compiles, but breaks the code. GCC ends up relying upon eax
not being clobbered, which is incorrect:
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
我怎么做我想要什么?
How do I do what I want?
推荐答案
只要为输出进入,编译器将优化出来一个无用的温度。例如:
Just make a useless temp for the output to go into and the compiler will optimize it out. For example:
__asm__("mull\t%2" : "=d"(div10), "=a"((int){0})
: "a"(UINT32_C(0x1999999A)), "r"(number) : "cc");
这是我所知道的处理被破坏输入的最简单的方法。
That's the easiest way I know to handle clobbered inputs.
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