std::numeric_limits<T>::signaling_NaN
来自cppreference.com
< cpp | types | numeric limits
| static T signaling_NaN() throw(); |
(C++11 前) | |
| static constexpr T signaling_NaN() noexcept; |
(C++11 起) | |
返回特殊值“发信的非数”，以浮点类型 T 表示。仅若 std::numeric_limits<T>::has_signaling_NaN == true 才有意义。在最常见的浮点数二进制表示 IEEE 754 中，任何所有指数位均为 1 且至少有一位尾数位为 1 的值表示 NaN 。何种尾数位的值表示安静或发信的 NaN ，及符号位是否有意义，是实现定义的。
返回值
T
|
std::numeric_limits<T>::signaling_NaN() |
| /* non-specialized */ | T()
|
| bool | false |
| char | 0 |
| signed char | 0 |
| unsigned char | 0 |
| wchar_t | 0 |
| char8_t | 0 |
| char16_t | 0 |
| char32_t | 0 |
| short | 0 |
| unsigned short | 0 |
| int | 0 |
| unsigned int | 0 |
| long | 0 |
| unsigned long | 0 |
| long long | 0 |
| unsigned long long | 0 |
| float | 实现定义 |
| double | 实现定义 |
| long double | 实现定义 |
注意
NaN 决不与自身比较相等。 IEEE-754 不要求复制 NaN 保留其位表示（符号与载荷），尽管大多数实现都会保留。
当发信 NaN 被用于算术表达式的参数时，可能引发适当的浮点异常，而且 NaN 被“安静化”，即表达式返回安静的 NaN 。
示例
演示用发信的 NaN 引发浮点异常
运行此代码
#include <iostream> #include <limits> #include <cfenv> #pragma STDC_FENV_ACCESS on void show_fe_exceptions() { int n = std::fetestexcept(FE_ALL_EXCEPT); if(n & FE_INVALID) std::cout << "FE_INVALID is raised\n"; else if(n == 0) std::cout << "no exceptions are raised\n"; std::feclearexcept(FE_ALL_EXCEPT); } int main() { double snan = std::numeric_limits<double>::signaling_NaN(); std::cout << "After sNaN was obtained "; show_fe_exceptions(); double qnan = snan * 2.0; std::cout << "After sNaN was multiplied by 2 "; show_fe_exceptions(); double qnan2 = qnan * 2.0; std::cout << "After the quieted NaN was multiplied by 2 "; show_fe_exceptions(); std::cout << "The result is " << qnan2 << '\n'; }
输出：
After sNaN was obtained no exceptions are raised After sNaN was multiplied by 2 FE_INVALID is raised After the quieted NaN was multiplied by 2 no exceptions are raised The result is nan
参阅
| 鉴别能表示特殊值“发信的非数”（ NaN ）的浮点类型 (公开静态成员常量) | |
| [静态] |
返回给定浮点类型的安静 NaN 值 (公开静态成员函数) |
| (C++11) |
检查给定的数是否 NaN (函数) |
