带有const参数和重载的函数

问题描述

尝试了所以它让我想为什么不重载的函数，我想出了一个稍微不同的代码，但它说，该函数不能重载。我的问题是为什么？或者有另一种方式？

Was tryin out the stackeroverflow qn so it got me thinking why not overload the the function and I came up with a slightly different code but it says the function cannot be overloaded. My question is why? or is there a another way?

 #include <iostream>
 using std::cout;

 class Test {
         public:
         Test(){ }
         int foo (const int) const;
         int foo (int );
 };

 int main ()
 {
         Test obj;
         Test const obj1;
         int variable=0;
     do{
         obj.foo(3);        // Call the const function
          obj.foo(variable); // Want to make it call the non const function
         variable++;
             usleep (2000000);
        }while(1);
 }

 int Test::foo(int a)
 {
    cout<<"NON CONST"<<std::endl;
    a++;
    return a;
 }

 int Test::foo (const int a) const
 {
    cout<<"CONST"<<std::endl;
    return a;
 }

推荐答案

只对非指针的constness，非引用类型。

You can't overload based only on the constness of a non pointer, non reference type.

想想如果你是编译器。
面对这行：

Think for instance if you were the compiler.Faced with the line:

 cout <<obj.foo(3);

您可以调用哪个函数？

当你通过值时，值将被复制。参数上的const只与函数定义相关。

As you are passing by value the value gets copied either way. The const on the argument is only relevant to the function definition.

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