[英]What's the meaning of the following code?
There is a CAS code below which can handle just int type,I know the function of CAS but I don't know the details shown below. 下面有一个CAS代码,它只能处理int类型,我知道CAS的功能,但是我不知道下面显示的细节。
inline int CAS(unsigned long *mem,unsigned long newval,unsigned long oldval)
{
__typeof (*mem) ret;
__asm __volatile ("lock; cmpxchgl %2, %1"
: "=a" (ret), "=m" (*mem)
: "r" (newval), "m" (*mem), "0" (oldval));
return (int) ret;
}
I know there should be five parameters mapped to %0,%1,%2,%3,%4 because there are five parameters in input/output field 我知道应该有五个参数映射到%0,%1,%2,%3,%4,因为在输入/输出字段中有五个参数
I also know that "=a"
means using eax
register, "=m"
means using memory address, "r"
means using any register 我也知道
"=a"
表示使用eax
寄存器, "=m"
表示使用内存地址, "r"
表示使用任何寄存器
But I don't understand what the "0" means. 但是我不明白“ 0”的含义。
I don't understand why "cmpxchgl" only use two parameters %2, %1 instead of three? 我不明白为什么“ cmpxchgl”仅使用两个参数%2,%1而不是三个?
It should use three params as the CAS function. 它应该使用三个参数作为CAS函数。
Where can I get all the infimation about the inline c asm?I need a complete tutorial. 我在哪里可以获得有关内联c asm的所有信息?我需要完整的教程。
%2
is newval
, %1
is *mem
%2
是newval
, %1
是*mem
with "0" (oldval)
, and the first register occur is "=a"
, means that oldval
is stored in eax
. 为
"0" (oldval)
,并且第一个寄存器为"=a"
,表示oldval
存储在eax
。
So cmpxchgl %2, %1"
means cmpxchgl newval, *mem"
(while oldval
in eax
), which checks eax
(value of oldval) whether equals *mem
, if equal, change value of *mem
to newval
. 因此,
cmpxchgl %2, %1"
表示cmpxchgl newval, *mem"
(而eax
oldval
),它检查eax
(oldval的值)是否等于*mem
,如果相等,则将*mem
值更改为newval
。
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