本文介绍了带有 const 参数和重载的函数的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着跟版网的小编来一起学习吧！
问题描述
正在试用 stackeroverflow qn 所以它让我思考为什么不重载该函数，我想出了一个稍微不同的代码，但它说该函数不能被重载.我的问题是为什么?或者有其他方法吗?
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
不能仅基于非指针、非引用类型的常量性进行重载.
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.面对线:
Think for instance if you were the compiler. Faced with the line:
cout <<obj.foo(3);
你会调用哪个函数?
当您按值传递时，值会以任何方式复制.参数上的 const 仅与函数定义相关.
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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