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

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.)

其他回答

一般来说,在C/ c++中没有标准的signum函数,缺少这样一个基本函数说明了很多关于这些语言的信息。

除此之外,我相信关于定义这样一个函数的正确方法的两种主流观点在某种程度上是正确的,而且一旦你考虑到两个重要的警告,关于它的“争议”实际上是没有争议的:

A signum function should always return the type of its operand, similarly to an abs() function, because signum is usually used for multiplication with an absolute value after the latter has been processed somehow. Therefore, the major use case of signum is not comparisons but arithmetic, and the latter shouldn't involve any expensive integer-to/from-floating-point conversions. Floating point types do not feature a single exact zero value: +0.0 can be interpreted as "infinitesimally above zero", and -0.0 as "infinitesimally below zero". That's the reason why comparisons involving zero must internally check against both values, and an expression like x == 0.0 can be dangerous.

对于C语言,我认为使用整型的最佳方法确实是使用(x > 0) - (x < 0)表达式,因为它应该以一种无分支的方式进行转换,并且只需要三个基本操作。最好定义强制返回类型与实参类型匹配的内联函数,并添加C11 define _Generic来将这些函数映射到公共名称。

With floating point values, I think inline functions based on C11 copysignf(1.0f, x), copysign(1.0, x), and copysignl(1.0l, x) are the way to go, simply because they're also highly likely to be branch-free, and additionally do not require casting the result from integer back into a floating point value. You should probably comment prominently that your floating point implementations of signum will not return zero because of the peculiarities of floating point zero values, processing time considerations, and also because it is often very useful in floating point arithmetic to receive the correct -1/+1 sign, even for zero values.

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)

我不知道它的标准函数。这里有一种有趣的写法:

(x > 0) - (x < 0)

这里有一个更容易理解的方法:

if (x > 0) return 1;
if (x < 0) return -1;
return 0;

如果你喜欢三元运算符,你可以这样做:

(x > 0) ? 1 : ((x < 0) ? -1 : 0)

在C/ c++中是否存在标准符号函数(signum, sgn) ?

是的,这取决于定义。

C99及以后版本在<math.h>中有signbit()宏

Int符号位(实浮点x); signbit宏当且仅当其参数值的符号为负时返回一个非零值。C11§7.12.3.6


然而OP想要一些不同的东西。

我想要一个函数,返回-1的负数和+1的正数. ...处理浮点数的函数。

#define signbit_p1_or_n1(x)  ((signbit(x) ?  -1 : 1)

更深层次的问题:

OP的问题在以下情况下不是特定的:x = 0.0, -0.0, +NaN, -NaN。

经典的signum()在x>0时返回+1,在x<0时返回-1,在x==0时返回0。

许多答案已经涵盖了这一点,但没有涉及x = -0.0, +NaN, -NaN。许多都是针对通常缺少Not-a-Numbers (NaN)和-0.0的整数观点。

典型答案函数如signnum_typical()在-0.0,+NaN, -NaN上,它们返回0.0,0.0,0.0。

int signnum_typical(double x) {
  if (x > 0.0) return 1;
  if (x < 0.0) return -1;
  return 0;
}

相反,我建议这样的功能:在-0.0,+NaN, -NaN上,它返回-0.0,+NaN, -NaN。

double signnum_c(double x) {
  if (x > 0.0) return 1.0;
  if (x < 0.0) return -1.0;
  return x;
}

有一个C99数学库函数叫copysign(),它从一个参数取符号,从另一个参数取绝对值:

result = copysign(1.0, value) // double
result = copysignf(1.0, value) // float
result = copysignl(1.0, value) // long double

会给你一个+/- 1.0的结果,这取决于值的符号。注意,浮点零是有符号的:(+0)将产生+1,(-0)将产生-1。