无法将std :: unorded_set与自定义KeyEqual进行比较

问题描述

以下程序无法编译.但是，如果我不注释 operator == ，它就会编译.当我已经提供了 FooEqual

The following program does not compile. But If I do not comment out operator==, it compiles. Why operator== is still needed when I already provide FooEqual

#include <cstddef>
#include <unordered_set>

struct Foo {
};

struct FooHasher {
  size_t operator()(const Foo&) const {
    return 1;
  }
};

struct FooEqual {
  bool operator()(const Foo& lhs, const Foo& rhs) const {
    return true;
  }
};

// bool operator==(const Foo& lhs, const Foo& rhs) {
//   return true;
// }

int main() {
  std::unordered_set<Foo, FooHasher, FooEqual> s1;
  std::unordered_set<Foo, FooHasher, FooEqual> s2;
  (void)(s1 == s2);
  return 0;
}

推荐答案

"23.2.5无序关联容器"状态:

"23.2.5 Unordered associative containers" states:

将其剥离，归结为根据 std :: is_permutation()定义的无序容器的相等性.

Stripping this down, it all comes down to the equality of unordered containers being defined in terms of std::is_permutation().

重要的是，它引用了 std :: is_permutation ()的三个参数形式，而不是四个参数形式！

The important part is that this references the three argument form of std::is_permutation(), and not the four argument form!

换句话说，对于无序容器的内容，而不是容器的官方比较功能，整个纸牌屋最终被缩减为默认的 operator == .

In other words, the whole house of cards ends up being reduced to the default operator==, for the contents of the unordered container, rather than the container's official comparison function.

这是我的读物.

这篇关于无法将std :: unorded_set与自定义KeyEqual进行比较的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！