通过引用与通过指针传递? [英] Pass by reference vs pass by pointer?

问题描述

可能重复:
何时通过引用和何时在C ++中通过指针传递?

通过引用传递和通过指针传递值之间有什么区别?

What is the difference between passing by reference and passing the value by a pointer?

推荐答案

通过引用传递参数时，函数内部的参数是您从外部传递的变量的别名.当您通过指针传递变量时，您将获取变量的地址并将该地址传递给函数.主要区别在于，您可以将不带地址的值(如数字)传递给采用const引用的函数，而不能将不带地址的值传递给采用const指针的函数.

When you pass a parameter by reference, the parameter inside the function is an alias to the variable you passed from the outside. When you pass a variable by a pointer, you take the address of the variable and pass the address into the function. The main difference is that you can pass values without an address (like a number) into a function which takes a const reference, while you can't pass address-less values into a function which takes const pointers.

通常，C ++编译器将引用实现为隐藏指针.

Typically a C++ compiler implement a reference as a hidden pointer.

您可以通过以下方式将函数更改为指针变量:

You can change your function into the pointer variant this way:

void flip(int *i) // change the parameter to a pointer type
{
    cout << "          flip start "<<"i="<< *i<<"\n"; // replace i by *i
    *i = 2*(*i); // I'm not sure it the parenthesis is really needed here,
                 // but IMHO this is better readable
    cout << "          flip exit  "<<"i="<< *i<<"\n";
}

int main()
{
    int j =1;
    cout <<"main j="<<j<<endl;
    flip(&j); // take the address of j and pass this value
    // adjust all other references ...
}

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