C语言 C中整数的快速符号
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Fast sign of integer in C
提问by Alex
There is a sign function in C:
C中有一个符号函数：
int sign(int x)
{
if(x > 0) return 1;
if(x < 0) return -1;
return 0;
}
Unfortunately, comparison cost is very high, so I need to modify function in order reduce the number of comparisons.
不幸的是，比较成本非常高，所以我需要修改函数以减少比较次数。
I tried the following:
我尝试了以下方法：
int sign(int x)
{
int result;
result = (-1)*(((unsigned int)x)>>31);
if (x > 0) return 1;
return result;
}
In this case I get only one comparison.
在这种情况下，我只得到一个比较。
Is there any way to avoid comparisons at all?
有没有办法完全避免比较？
EDITpossible duplicatedoes not give an answer for a question as all answers are C++, uses comparison (that I supposed to avoid) or does not return -1, +1, 0.
编辑可能的重复不会给出问题的答案，因为所有答案都是 C++，使用比较（我应该避免）或不返回-1, +1, 0。
回答by NPE
First of all, integer comparison is very cheap. It's branchingthat can be expensive (due to the risk of branch mispredictions).
首先，整数比较非常便宜。它的分支，可以是昂贵的（由于分支预测失误的风险）。
I have benchmarked your function on a Sandy Bridge box using gcc 4.7.2, and it takes about 1.2ns per call.
我已经使用 gcc 4.7.2 在 Sandy Bridge 机器上对您的函数进行了基准测试，每次调用大约需要 1.2ns。
The following is about 25% faster, at about 0.9ns per call:
以下大约快 25%，每次调用大约 0.9ns：
int sign(int x) {
return (x > 0) - (x < 0);
}
The machine code for the above is completely branchless:
上面的机器代码是完全无分支的：
_sign:
xorl %eax, %eax
testl %edi, %edi
setg %al
shrl , %edi
subl %edi, %eax
ret
Two things are worth pointing out:
有两点值得指出：
- The base level of performance is very high.
- Eliminating branching does improve performance here, but not dramatically.
- 性能的基础水平非常高。
- 消除分支确实可以提高这里的性能，但不会显着提高。
回答by Chris Dodd
int sign(int x)
{
// assumes 32-bit int and 2s complement signed shifts work (implementation defined by C spec)
return (x>>31) | ((unsigned)-x >> 31);
}
The first part (x>>32) gives you -1 for negative numbers and 0 for 0 or positive numbers.
The second part gives you 1 if x > 0 or equal to INT_MIN, and 0 otherwise. Or gives you the right final answer.
第一部分 ( x>>32) 为负数提供 -1，0 或正数为 0。如果 x > 0 或等于 INT_MIN，则第二部分为 1，否则为 0。或者给你正确的最终答案。
There's also the canonical return (x > 0) - (x < 0);, but unfortunately most compilers will use branches to generate code for that, even though there are no visible branches. You can try to manually turn it into branchless code as:
还有 canonical return (x > 0) - (x < 0);，但不幸的是，大多数编译器会使用分支来为此生成代码，即使没有可见的分支。您可以尝试手动将其转换为无分支代码：
int sign(int x)
{
// assumes 32-bit int/unsigned
return ((unsigned)-x >> 31) - ((unsigned)x >> 31);
}
which is arguably better than the above as it doesn't depend on implementation defined behavior, but has a subtle bug in that it will return 0 for INT_MIN.
这可以说比上面的更好，因为它不依赖于实现定义的行为，但有一个微妙的错误，它会为 INT_MIN 返回 0。
回答by Bruce
int sign(int x) {
return (x>>31)|(!!x);
}
回答by anatolyg
If s(x)is a function that returns the sign-bit of x(you implemented it by ((unsigned int)x)>>31), you can combine s(x)and s(-x)in some way. Here is a "truth table":
Ifs(x)是一个返回符号位的函数x（您通过 实现了它((unsigned int)x)>>31），您可以以某种方式组合s(x)和s(-x)。这是一个“真值表”：
x > 0: s(x) = 0; s(-x) = 1; your function must return 1
x < 0: s(x) = 1; s(-x) = 0; your function must return -1
x = 0: s(x) = 0; s(-x) = 0; your function must return 0
x > 0: s(x) = 0; s(-x) = 1; 你的函数必须返回 1
x < 0: s(x) = 1; s(-x) = 0; 你的函数必须返回 -1
x = 0: s(x) = 0; s(-x) = 0; 你的函数必须返回 0
So you can combine them in the following way:
因此，您可以通过以下方式组合它们：
s(-x) - s(x)
回答by Aniket Inge
int i = -10;
if((i & 1 << 31) == 0x80000000)sign = 0;else sign = 1;
//sign 1 = -ve, sign 0 = -ve
