在 C++ 中通过引用传递对象

问题描述

在 C++(也包括 C)中通过引用传递变量的常用方法如下:

The usual way to pass a variable by reference in C++(also C) is as follows:

void _someFunction(dataType *name){ // dataType e.g int,char,float etc.
/****
definition
*/
}

int main(){
    dataType v;
    _somefunction(&v);  //address of variable v being passed
    return 0;
}

但令我惊讶的是，我注意到当通过引用传递一个对象时，对象的名称本身就起到了作用(不需要 & 符号)，并且在声明/函数定义 参数前不需要*符号.下面的例子应该很清楚:

But to my surprise, I noticed when passing an object by reference the name of object itself serves the purpose(no & symbol required) and that during declaration/definition of function no * symbol is required before the argument. The following example should make it clear:

// this
#include <iostream>
using namespace std;

class CDummy {
  public:
    int isitme (CDummy& param);     //why not (CDummy* param);
};

int CDummy::isitme (CDummy& param)
{
  if (&param == this) return true;
  else return false;
}

int main () {
  CDummy a;
  CDummy* b = &a;
  if ( b->isitme(a) )               //why not isitme(&a)
    cout << "yes, &a is b";
  return 0;
}

我无法理解为什么要对 class 进行这种特殊处理.即使是几乎像一个类的结构也不是这样使用的.对象名称是否和数组一样被视为地址?

I have problem understanding why is this special treatment done with class . Even structures which are almost like a class are not used this way. Is object name treated as address as in case of arrays?

推荐答案

让您感到困惑的是，声明为传递引用的函数(使用 &) 不使用实际地址调用，即 &a.

What seems to be confusing you is the fact that functions that are declared to be pass-by-reference (using the &) aren't called using actual addresses, i.e. &a.

简单的答案是将函数声明为传递引用:

The simple answer is that declaring a function as pass-by-reference:

void foo(int& x);

就是我们所需要的.然后它会自动通过引用传递.

is all we need. It's then passed by reference automatically.

你现在像这样调用这个函数:

You now call this function like so:

int y = 5;
foo(y);

和 y 将通过引用传递.

and y will be passed by reference.

你也可以这样做(但为什么要这样做?咒语是:尽可能使用引用，需要时使用指针):

You could also do it like this (but why would you? The mantra is: Use references when possible, pointers when needed) :

#include <iostream>
using namespace std;

class CDummy {
public:
    int isitme (CDummy* param);
};


int CDummy::isitme (CDummy* param)
{
    if (param == this) return true;
    else return false;
}

int main () {
    CDummy a;
    CDummy* b = &a;             // assigning address of a to b
    if ( b->isitme(&a) )        // Called with &a (address of a) instead of a
        cout << "yes, &a is b";
    return 0;
}

输出:

yes, &a is b

这篇关于在 C++ 中通过引用传递对象的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！