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C++ 中的运算符重载

发布于 2024-08-04 18:36:20 字数 78 浏览 4 评论 0原文

除了“新建”、“删除”、“<<” & '>>'运算符，还有哪些其他运算符可以在类上下文之外的 C++ 中重载？

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墨离汐 2024-08-11 18:36:20

以下运算符（由空格分隔）可以作为非成员函数重载：

new delete new[] delete[] + - * / % ˆ & | ˜ ! < > += -= *= /= %= ˆ=  
&= |= << >> >>= <<= == != <= >= && || ++ -- , ->*

以下运算符必须是非静态成员函数：

-> () [] =

以下运算符不能重载：

. .* :: ?: # ##

转换运算符也必须是成员函数。

仅仅因为它有一个“=”并不意味着它不能作为非成员运算符重载。以下是格式良好的：

struct A { };
A operator +=(A,A) { return A(); }
A a = A()+=A();

并且前缀和后缀递增和递减运算符确实可以定义为非成员：

13.5.7 名为operator++ 的用户定义函数实现了前缀和后缀++ 运算符。如果该函数是不带参数的成员函数，或者是具有一个类或枚举类型参数的非成员函数，则它为该类型的对象定义前缀增量运算符 ++。如果该函数是具有一个参数的成员函数（应为 int 类型）或具有两个参数的非成员函数（其中第二个参数应为 int 类型），则它为以下对象定义后缀增量运算符 ++那种类型。当由于使用 ++ 运算符而调用后缀增量时，int
参数的值为零。
前缀和后缀递减运算符 -- 处理方式类似

标准中的第 13.5 条涵盖了这一点。

希望这有帮助。

The following operators (delimitted by space) can be overloaded as non-member functions:

new delete new[] delete[] + - * / % ˆ & | ˜ ! < > += -= *= /= %= ˆ=  
&= |= << >> >>= <<= == != <= >= && || ++ -- , ->*

The following have to be non-static member functions:

-> () [] =

The following can not be overloaded:

. .* :: ?: # ##

conversion operators also have to be member functions.

And just because it has a '=' in it does not mean it cannot be overloaded as a non-member operator. The following is well-formed:

struct A { };
A operator +=(A,A) { return A(); }
A a = A()+=A();

And the prefix and postfix increment and decrement operators can indeed be defined as non-members:

13.5.7 The user-defined function called operator++ implements the prefix and postfix ++ operator. If this function is a member function with no parameters, or a non-member function with one parameter of class or enumeration type, it defines the prefix increment operator ++ for objects of that type. If the function is a member function with one parameter (which shall be of type int) or a non-member function with two parameters (the second of which shall be of type int), it defines the postfix increment operator ++ for objects of that type. When the postfix increment is called as a result of using the ++ operator, the int
argument will have value zero.
The prefix and postfix decrement operators -- are handled analogously

Clause 13.5 in the Standard covers this.

Hope this helps.

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