为什么std :: min仅支持initializer_list? [英] Why does std::min only support initializer_list?

问题描述

我们可以通过以下方式使用 std :: min :

We can use std::min in below way:

// 1.
int a = 1, b = 2;
std::min(a, b);

// 2.
std::min({1,2,3,4});

但是为什么不能使用 std :: vector 或 std :: list ，因为模板中的参数是 initializer_list .

But why can't use a std::vector or std::list, Because the param in the template is initializer_list.

template <class T, class Compare>
  pair<T,T> minmax (initializer_list<T> il, Compare comp);

这种设计的原因是什么?

What is the reason for this design?

推荐答案

要解释为什么它不接受容器"，请考虑以下语义:

To explain "why it doesn't accept a container", take the semantic into consideration:

std :: min({"foo"，"bar"，"hello"})

std :: min()的语义表示在输入参数中找到最小值".因此， std :: min()/ std :: max()接受两个参数，或者将initializer_list用作更多参数".

The semantic of std::min() means "find the minimum value in the input parameters". Thus std::min()/std::max() takes two arguments, or an initializer_list as "more arguments".

std :: min()无法提供通过容器迭代"的功能，因为容器被视为参数".

std::min() doesn't provide the ability to "iterate through a container", because a container is considered as "a parameter".

要查找容器中的最小值，请在C +中找到 std :: min_element()和eerorika的建议 std :: ranges :: min()+20应该更好.

To find the min value in a container, there is std::min_element(), and eerorika's suggestion std::ranges::min() in C++20 should be better.

对于 std :: min_element()的用法，您可以参考如何获取最大值(或min)值中的值?.

For std::min_element() usage, you may refer to How can I get the max (or min) value in a vector?.

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