从另一个非常相似的问题(dupe?)中窃取一个c++特定的答案:

函数返回类型不会在重载解析中发挥作用，因为Stroustrup(我假设来自其他c++架构师的输入)希望重载解析是“上下文独立的”。参见“c++编程语言，第三版”中的“重载和返回类型”。

原因是为了保持独立于上下文的单个操作符或函数调用的解析。

They wanted it to be based only on how the overload was called - not how the result was used (if it was used at all). Indeed, many functions are called without using the result or the result would be used as part of a larger expression. One factor that I'm sure came into play when they decided this was that if the return type was part of the resolution there would be many calls to overloaded functions that would need to be resolved with complex rules or would have to have the compiler throw an error that the call was ambiguous.

而且，上帝知道，c++的重载解析已经足够复杂了……

如果函数被返回类型重载并且有这两个重载

int func();
string func();

在看到这样的调用时，编译器无法确定调用这两个函数中的哪一个

void main() 
{
    func();
}

由于这个原因，语言设计者通常不允许返回值重载。

然而，有些语言(如MSIL)允许按返回类型重载。当然，它们也面临上述困难，但它们有变通办法，为此您必须查阅它们的文档。

如果你想重载具有不同返回类型的方法，只需添加一个具有默认值的虚拟参数来允许重载执行，但不要忘记参数类型应该是不同的，因此重载逻辑工作接下来是delphi上的示例:

type    
    myclass = class
    public
      function Funct1(dummy: string = EmptyStr): String; overload;
      function Funct1(dummy: Integer = -1): Integer; overload;
    end;

像这样使用它

procedure tester;
var yourobject : myclass;
  iValue: integer;
  sValue: string;
begin
  yourobject:= myclass.create;
  iValue:= yourobject.Funct1(); //this will call the func with integer result
  sValue:= yourobject.Funct1(); //this will call the func with string result
end;

从另一个非常相似的问题(dupe?)中窃取一个c++特定的答案:

函数返回类型不会在重载解析中发挥作用，因为Stroustrup(我假设来自其他c++架构师的输入)希望重载解析是“上下文独立的”。参见“c++编程语言，第三版”中的“重载和返回类型”。

原因是为了保持独立于上下文的单个操作符或函数调用的解析。

They wanted it to be based only on how the overload was called - not how the result was used (if it was used at all). Indeed, many functions are called without using the result or the result would be used as part of a larger expression. One factor that I'm sure came into play when they decided this was that if the return type was part of the resolution there would be many calls to overloaded functions that would need to be resolved with complex rules or would have to have the compiler throw an error that the call was ambiguous.

而且，上帝知道，c++的重载解析已经足够复杂了……

如前所述，对仅因返回类型不同而不同的函数的模糊调用会引入模糊。 模糊性会导致有缺陷的代码。 必须避免有缺陷的代码。

试图模糊化所带来的复杂性表明这不是一个好的hack。 除了智力练习之外，为什么不使用带有引用参数的过程呢?

procedure(reference string){};
procedure(reference int){};
string blah;
procedure(blah)

我认为这是现代c++定义中的一个GAP……为什么?

int func();
double func();

// example 1. → defined
int i = func();

// example 2. → defined
double d = func();

// example 3. → NOT defined. error
void main() 
{
    func();
}

为什么c++编译器不能抛出例子“3”和错误 接受例子“1+2”中的代码??