自动从派生类转换为基类成员变量的类型

问题描述

长话短说：我想理解为什么在下面的代码的最后一行中没有使用 D :: operator B（）const 转换运算符，因此，当使用 g ++ -std = c ++ 17 source.cpp 编译时（使用 g ++ -std = c ++ 2a deleteme.cpp编译时失败） 成功）。

Long story short: I'd like to understand why the D::operator B() const conversion operator is not used in the last line in the code below, which thus fails when compiling with g++ -std=c++17 source.cpp (compiling with g++ -std=c++2a deleteme.cpp is successful, though).

错误是：

$ g++ -std=c++17 deleteme.cpp && ./a.out
In file included from /usr/include/c++/10.2.0/cassert:44,
                 from deleteme.cpp:1:
deleteme.cpp: In function ‘int main()’:
deleteme.cpp:19:14: error: no match for ‘operator==’ (operand types are ‘D’ and ‘B’)
   19 |     assert(d == B{2}); // conversion operator not invoked explicitly errors // LINE
      |            ~ ^~ ~~~~
      |            |    |
      |            D    B
In file included from /usr/include/c++/10.2.0/utility:70,
                 from deleteme.cpp:2:
/usr/include/c++/10.2.0/bits/stl_pair.h:466:5: note: candidate: ‘template<class _T1, class _T2> constexpr bool std::operator==(const std::pair<_T1, _T2>&, const std::pair<_T1, _T2>&)’
  466 |     operator==(const pair<_T1, _T2>& __x, const pair<_T1, _T2>& __y)
      |     ^~~~~~~~
/usr/include/c++/10.2.0/bits/stl_pair.h:466:5: note:   template argument deduction/substitution failed:
In file included from /usr/include/c++/10.2.0/cassert:44,
                 from deleteme.cpp:1:
deleteme.cpp:19:20: note:   ‘B’ is not derived from ‘const std::pair<_T1, _T2>’
   19 |     assert(d == B{2}); // conversion operator not invoked explicitly errors // LINE
      |

代码为：

#include <cassert>
#include <utility>

struct B {
    int x;
    B(int x) : x(x) {}
    bool operator==(B const& other) const { return x == other.x; }
};

struct D : std::pair<B,char*> {
    operator B() const { return this->first; }
};

int main() {
    B b{1};
    D d{std::pair<B,char*>(B{2},(char*)"hello")};

    assert((B)d == B{2}); // conversion operator invoked explicitly is fine
    assert(d == B{2}); // conversion operator not invoked explicitly errors // LINE
}

此问题是对。在这里，我得到了帮助，编写了一个类 Recursive ，该类的行为类似于一对（因此继承自该类），其第一个是 std :: array ，其第二个是 boost :: hana :: optional< std :: vector< ...>> （有关详细信息，请参阅链接）。

This question is a follow up to this. There I got help to write a class Recursive which behaves like a pair (so inherits from it) whose first is a std::array and whose second is a boost::hana::optional<std::vector<...>> (see the link for details).

由于 std的秒： ：pair 有点高级，信息到 first 中包含的内容，在很多地方，我想强制转换/转换 std :: pair 的类 Recursive 为其 first 的类型。

Since the second of the std::pair is kind of an "advanced" information to what's contained in the first, in many places I'd like to cast/convert this object of std::pair-like class Recursive to the type of its first.

推荐答案

当编译器看到 d == B {2} 时，它首先

When compiler sees d == B{2}, it first creates a list of operator== overloads that it's able to find, and then performs overload resolution on them.

如链接所述，过载列表包含：

As the link explains, the overload list contains:

• 第一个操作数的成员 operator == s。


• 非成员 operators == s通过不合格的查找找到（如果有的话）。


• 内置 operator == s（如果您的操作数）可以转换为内置类型。

• Member operator==s of the first operand, if any.
• Non-member operators==s found by unqualified lookup, if any.
• Built-in operator==s, if your operands can be converted to built-in types.

没有提到检查第一个操作数s的转换运算符o您的 operator == 找不到。

There's no mention of examining conversion operators of the first operand, so your operator== doesn't get found.

解决方案是使 operator == 非成员（可能将其定义为朋友）。可行：

The solution is to make the operator== non-member (possibly define it as a friend). This works:

friend bool operator==(const B &a, const B &b) {return a.x == b.x;}




---

从C ++ 20开始，比较运算符的规则放松了：编译器将在第二个操作数中也查找成员 operator == ，并愉快地调用非成员 operator ==

---

Starting from C++20, the rules of comparison operators got relaxed: the compiler will look for member operator== in the second operand as well, and will happily call non-member operator== with arguments in a wrong order.

所以 a == b 和 b == a 现在等同于学位，除了：

So a == b and b == a are now equivalent to a degree, except that:

• 调用了 operator == 这种新方式必须返回 bool 。


• 如果给出选择，编译器将倾向于使用旧的调用的方式。调用rhs的成员 operator == 来代替operator == ，而不是调用非成员 operator == 参数顺序错误。

• An operator== called in this new manner must return bool.
• If given choice, the compiler will prefer the old ways of calling operator== over calling the member operator== of the rhs, which is in turn preferred over calling a non-member operator== with a wrong argument order.

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