在C ++中通过引用传递对象 [英] passing object by reference in C++

问题描述

在C ++（也是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，我注意到通过引用传递对象对象本身的名称用于目的（不需要& 符号）以及在声明/定义函数期间 * 符号在参数之前是必需的。
以下示例应该清楚：

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;
}

我有问题了解为什么这个特殊的处理类。即使是几乎像类的结构也不会这样使用。 对象名称是否像数组一样被视为地址？

推荐答案

是声明为通过引用传递的函数（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.

简单的答案是声明一个函数作为传递引用：

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

void foo(int& x);

是我们需要的。

您现在可以这样调用此函数：

You now call this function like so:

int y = 5;
foo(y);

和 y 将通过引用传递。

您也可以这样做（但是为什么？）咒语是：尽可能使用引用，需要时指针 / p>

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

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