std::size 的用户重载？

Question

在 std 命名空间中添加 size 的重载是否符合标准？我有一个无法更改的 class，但我希望 std::size 也能为它工作；在通用设置中。

Answer 1

It is undefined behavior to add declarations or definitions to namespace std or to any namespace nested within std, with a few exceptions noted below

在同一页的更下方（直到 C++20）：

It is allowed to add template specializations for any standard library function template to the namespace std only if the declaration depends on at least one program-defined type and the specialization satisfies all requirements for the original template, except where such specializations are prohibited.

但是从 C++20 开始：

It is undefined behavior to declare a full specialization of any standard library function template.

但是，由于 ADL，您无需为 std::size 提供专业化。这有效：

#include <array>
#include <iostream>
#include <cstddef>

template <typename T>
size_t get_size_example(const T& t) {
    using std::size;
    return size(t);
}

namespace Foo {
    struct bar {};
    size_t size(const bar& b){ return 42;}
}


int main(){
    std::array<int,5> x;
    Foo::bar b;
    std::cout << size(b) << "\n";
    std::cout << size(x) << "\n";
    std::cout << get_size_example(b) << "\n";
    std::cout << get_size_example(x) << "\n";
}

并打印：

std::size 的用户重载？

User overloads of std::size?

c++

c++-standard-library