问题
There is a sign function in C:
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.
I tried the following:
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.
Is there any way to avoid comparisons at all?
EDIT possible duplicate does 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
.
回答1:
First of all, integer comparison is very cheap. It's branching that can be expensive (due to the risk of branch mispredictions).
I have benchmarked your function on a Sandy Bridge box using gcc 4.7.2, and it takes about 1.2ns per call.
The following is about 25% faster, at about 0.9ns per call:
int sign(int x) {
return (x > 0) - (x < 0);
}
The machine code for the above is completely branchless:
_sign:
xorl %eax, %eax
testl %edi, %edi
setg %al
shrl $31, %edi
subl %edi, %eax
ret
Two things are worth pointing out:
- The base level of performance is very high.
- Eliminating branching does improve performance here, but not dramatically.
回答2:
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.
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:
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.
回答3:
int sign(int x) {
return (x>>31)|(!!x);
}
回答4:
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":
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
So you can combine them in the following way:
s(-x) - s(x)
回答5:
int i = -10;
if((i & 1 << 31) == 0x80000000)sign = 0;else sign = 1;
//sign 1 = -ve, sign 0 = -ve
来源:https://stackoverflow.com/questions/14579920/fast-sign-of-integer-in-c