英文:
operator== on C++ structs with conversion operators
问题
我试图理解当比较两个没有定义==操作符的结构体时所涉及的==操作符集合。在这种情况下，会使用int的内置操作符进行转换。然而，对于嵌套类来说，例如，如果Y包含一个定义了==操作符的类，这种方法就不起作用。这只适用于结构体可以无歧义地转换为内置类型的情况吗？
英文:
I'm trying to understand the set of == operators that are explored when comparing two structs where there is no operator == defined. In this case the built-in operator for ints is used for conversion. However this does not work for nested classes for example, i.e. if Y contained a class that defined an == operator. Is this only applicable when a struct can be converted to a built-in type unambiguously?
struct X {
int x;
};
struct Y {
int y;
operator int() {
return y;
}
};
int main() {
// bool val {X{} == X{}}; // does not compile
bool val {Y{} == Y{}}; // works
}
答案1
得分: 2
以下是翻译好的部分：
类和结构体默认没有等于（或其他）运算符。您需要为您的实现重载所需的运算符并明确指定比较方式。
struct X {
int x;
bool operator==(const X& other) const { return x == other.x; }
bool operator!=(const X& other) const { return y != other.x; }
...
};
对于每个比较运算符（==、!=、<、>、<=、>=），都是如此，除非您使用C++20，在C++20中，您可以指定默认的<=>运算符，并让编译器为您生成它们（当然，您的结构体成员越多，这个过程就越复杂；关于比较的文档是很好的阅读材料）。
#include <compare>
struct X {
int x;
auto operator<=>(const X&) const = default;
};
英文:
Classes and structs don't have equality (or any) operators by default. You need to overload whatever is needed for your implementation and specify the comparisons explicitly.
struct X {
int x;
bool operator==(const X& other) const { return x == other.x; }
bool operator!=(const X& other) const { return y != other.x; }
...
};
That's true for every comparison operator (==, !=, <, >, <=, >=). That is unless you use C++20, in which you can specify the default <=> operator and have the compiler generate them for you (ofc the more members your struct has, the more complicated this gets; the docs on comparisons are a good read).
#include <compare>
struct X {
int x;
auto operator<=>(const X&) const = default;
};
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