重载转换运算符的后果

Consequences of overloading the conversion operator

我有一个 class 有一个像这样的重载转换运算符:

template <class T> class Pointer {
    T* object;
public:
    Pointer (T* object): object(object) {}
    operator T* () { return object; }
};

我注意到一些我通常必须手动重载的运算符现在突然像 Pointer 一样工作 T* 但有些运算符不是:

MyClass my_object;
Pointer<MyClass> pointer (&my_object);
if (pointer) { /* ... */ } // works
if (pointer == &my_object) { /* ... */ } // works
(*pointer).some_method (); // works
pointer->some_method (); // doesn't work
pointer = pointer + 1; // works
pointer++; // doesn't work

假设这是符合标准的正确行为,我如何知道哪些有效,哪些无效(无需反复试验),更重要的是,为什么会这样?

上面的一些操作之所以有效,是因为编译器可以使用 SINGLE 用户定义的转换隐式地将自定义类型:Pointer<MyClass> 转换为原始指针:MyClass*。 隐式转换有严格的规则,如 here.

所述

Implicit conversions are performed whenever an expression of some type T1 is used in context that does not accept that type, but accepts some other type T2, in particular:

(1) When the expression is used as the argument when calling a function that is declared with T2 as parameter.
(2) When the expression is used as an operand with an operator that expects T2
(3) When initializing a new object of type T2, including return statement in a function returning T2.
(4) When the expression is used in a switch statement (T2 is integral type)
(5) When the expression is used in an if statement or a loop (T2 is bool)

根据上面的例子进行一些调查,如果有人可以验证或更正我的推论,我将不胜感激。

  • 案例(5),if语句:

    if (pointer) { /* ... */ } // works

  • case (2), operator== with operand which can be implicitly converted to MyClass*:

    if (pointer == &my_object) { /* ... */ } // works

  • 情况(2),间接(operator*)取操作数可以隐式转换为MyClass*,然后结构引用(operator.):

    (*pointer).some_method (); // works

  • 不匹配任何情况,operator->不接受任何参数,可以隐式转换:

    pointer->some_method (); // doesn't work

  • case (2): operator+ 获取可以隐式转换为 MyClass* 的操作数,并使用构造函数将 return 值赋给 Pointer<MyClass>operator=;请注意,将 explicit 添加到构造函数会阻止编译,因为 return 类型的表达式:pointer + 1MyClass*,因此隐式调用采用 MyClass* 的构造函数:

    pointer = pointer + 1; // works

  • 不符合任何一种情况;请注意,即使显式转换为 MyClass*(例如 static_cast<MyClass*>(pointer)++)也无​​济于事,因为这里需要左值;解决方法是:auto ptr = &(*pointer); ptr++;:

    pointer++; // doesn't work

请记住,重载转换运算符有时会导致危险情况,例如。 MyClass* ptr = pointer; delete ptr; 将删除基础资源,编译器甚至不会抱怨。