我想要一个函数为负数返回-1,为正数返回+1。 http://en.wikipedia.org/wiki/Sign_function 我自己编写它很容易,但它似乎应该存在于某个标准库中。
编辑:具体来说,我正在寻找一个函数工作在浮动。
我想要一个函数为负数返回-1,为正数返回+1。 http://en.wikipedia.org/wiki/Sign_function 我自己编写它很容易,但它似乎应该存在于某个标准库中。
编辑:具体来说,我正在寻找一个函数工作在浮动。
当前回答
不,它在c++中不存在,就像在matlab中一样。我在程序中使用宏来实现这一点。
#define sign(a) ( ( (a) < 0 ) ? -1 : ( (a) > 0 ) )
其他回答
比上述解决方案更快,包括评级最高的一个:
(x < 0) ? -1 : (x > 0)
double signof(double a) { return (a == 0) ? 0 : (a<0 ? -1 : 1); }
类型安全的c++版本:
template <typename T> int sgn(T val) {
return (T(0) < val) - (val < T(0));
}
好处:
Actually implements signum (-1, 0, or 1). Implementations here using copysign only return -1 or 1, which is not signum. Also, some implementations here are returning a float (or T) rather than an int, which seems wasteful. Works for ints, floats, doubles, unsigned shorts, or any custom types constructible from integer 0 and orderable. Fast! copysign is slow, especially if you need to promote and then narrow again. This is branchless and optimizes excellently Standards-compliant! The bitshift hack is neat, but only works for some bit representations, and doesn't work when you have an unsigned type. It could be provided as a manual specialization when appropriate. Accurate! Simple comparisons with zero can maintain the machine's internal high-precision representation (e.g. 80 bit on x87), and avoid a premature round to zero.
警告:
It's a template so it might take longer to compile in some circumstances. Apparently some people think use of a new, somewhat esoteric, and very slow standard library function that doesn't even really implement signum is more understandable. The < 0 part of the check triggers GCC's -Wtype-limits warning when instantiated for an unsigned type. You can avoid this by using some overloads: template <typename T> inline constexpr int signum(T x, std::false_type is_signed) { return T(0) < x; } template <typename T> inline constexpr int signum(T x, std::true_type is_signed) { return (T(0) < x) - (x < T(0)); } template <typename T> inline constexpr int signum(T x) { return signum(x, std::is_signed<T>()); } (Which is a good example of the first caveat.)
为什么使用三元操作符和if-else,当你可以简单地这样做
#define sgn(x) x==0 ? 0 : x/abs(x)
int sign(float n)
{
union { float f; std::uint32_t i; } u { n };
return 1 - ((u.i >> 31) << 1);
}
这个函数假设:
浮点数的二进制32表示 在使用命名联合时对严格的混叠规则做出例外的编译器