转换运算符如何返回值？ [英] How does conversion operator return a value?

问题描述

对于类A，整数转换操作符看起来像：

  operator int 't指定任何返回类型
 {
 return intValue; 
} 
  

上述函数如何在返回值类型出现时返回一个值不指定？它似乎没有返回任何东西，但我知道它不是 void 。

c> 运算符T（） c $ c>总是 T 。

它不使用标准函数原型语法 T foo（），因为2个函数（例如 int foo（）与冲突double foo（） ）。如果使用此语法，那么您只能定义1个转换运算符重载，这是不希望的。

For a class A, an integer conversion operator would look something like;

operator int() //Here we don't specify any return type
{
    return intValue;
}

How is the above function able to return a value when its return value type appears not to be specified? It doesn't appear to return "anything", but I know it's not void.

How is this meaningful when a return type is not specified?

解决方案

The return type of operator T() is always T. It's a special case of C++.

It does not use standard function prototype syntax T foo() because 2 functions with the same name differing only by the return type cannot coexist (e.g. int foo() conflicts with double foo()). If this syntax is used then you can only define 1 conversion operator overload, which is undesirable.

这篇关于转换运算符如何返回值？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IT屋！