为什么不返回一个整数?(复制)

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Why do round() and ceil() not return an integer? 2 answers
为什么round()和ceil()不返回整数?2答案

Just now I stumbled upon the fact, that the C++ function floor returns the same type you pass to it, be it float, double or such.

刚才我无意中发现了一个事实，即c++函数层返回的类型与您传递给它的类型相同，可以是float类型，double类型或此类类型。

According to this reference, the function returns a down rounded integral value. Why isn't this an integer?

根据此引用，函数返回一个向下的整数整数值。为什么这不是一个整数?

1 个解决方案

#1

Because an integral type can't necessarily hold the same integral values as a float or double.

因为整数类型不能与浮点数或双精度数保持相同的整数值。

int main(int argc, char *argv[]) {
    std::cout << floor(std::numeric_limits<float>::max()) << std::endl;
    std::cout << static_cast<long>(floor(std::numeric_limits<float>::max())) << ::endl;
}

outputs (on my x86_64 architecture)

输出(在我的x86_64架构上)

3.40282e+38
-9223372036854775808

Additionally, floating-point values can hold NaN, +Inf, and -Inf, all of which are preserved by a floor() operation. None of these values can be represented with an integral type.

此外，浮点值可以容纳NaN、+Inf和-Inf，所有这些都由一个层()操作保存。这些值都不能用整数类型表示。

int main(int argc, char *argv[]) {
    std::cout << floor(std::numeric_limits<float>::quiet_NaN()) << std::endl;
    std::cout << floor(std::numeric_limits<float>::infinity()) << std::endl;
    std::cout << floor(-std::numeric_limits<float>::infinity()) << std::endl;
}

outputs

输出

nan
inf
-inf

#1