問題描述

正在試用 stackeroverflow qn 所以它讓我思考為什么不重載該函數(shù)，我想出了一個(gè)稍微不同的代碼，但它說該函數(shù)不能被重載.我的問題是為什么?或者有其他方法嗎?

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;
 }

推薦答案

不能僅基于非指針、非引用類型的常量性進(jìn)行重載.

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

想想如果你是編譯器.面對(duì)線:

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

 cout <<obj.foo(3);

你會(huì)調(diào)用哪個(gè)函數(shù)?

當(dāng)您按值傳遞時(shí)，值會(huì)以任何方式復(fù)制.參數(shù)上的 const 僅與函數(shù)定義相關(guān).

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

這篇關(guān)于帶有 const 參數(shù)和重載的函數(shù)的文章就介紹到這了，希望我們推薦的答案對(duì)大家有所幫助，也希望大家多多支持html5模板網(wǎng)！