为什么是numeric_limits< int> :: max（）> numeric_limits< int> :: infinity（）？ [英] Why is numeric_limits<int>::max() > numeric_limits<int>::infinity()?

问题描述

我在阅读在C ++中设置int为Infinity 。我理解，当一个需要真正的无穷大，一个应该使用 numeric_limits< float> :: infinity（）;我想其背后的理由是，通常整数类型没有指定用于表示像IEEE 754浮点数的特殊状态（例如， Inf t任务既不 - int & float used都留给实现）但仍然误导 max>无穷大。我试图理解标准中这个调用背后的理由。如果 infinity 对一个类型没有意义，那么它不应该被禁止，而不是有一个标志被检查其有效性？

解决方案

函数 numeric_limits< T> :: infinity（） $ c> T 为 numeric_limits< T> :: has_infinity 返回 true / p>

如果 T = int ，则返回 false 。所以比较没有意义，因为 numeric_limits< int> :: infinity（）不会返回任何有意义的值。 / p>

I was reading Setting an int to Infinity in C++. I understand that when one needs true infinity, one is supposed to use numeric_limits<float>::infinity(); I guess the rationale behind it is that usually integral types have no values designated for representing special states like NaN, Inf, etc. like IEEE 754 floats do (again C++ doesn't mandate neither - int & float used are left to the implementation); but still it's misleading that max > infinity for a given type. I'm trying to understand the rationale behind this call in the standard. If having infinity doesn't make sense for a type, then shouldn't it be disallowed instead of having a flag to be checked for its validity?

解决方案

The function numeric_limits<T>::infinity() makes sense for those T for which numeric_limits<T>::has_infinity returns true.

In case of T=int, it returns false. So that comparison doesn't make sense, because numeric_limits<int>::infinity() does not return any meaningful value to compare with.

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