問(wèn)題描述

在 C++(也包括 C)中通過(guò)引用傳遞變量的常用方法如下:

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;
}

但令我驚訝的是，我注意到當(dāng)通過(guò)引用傳遞一個(gè)對(duì)象時(shí)，對(duì)象的名稱(chēng)本身就起到了作用(不需要 & 符號(hào))，并且在聲明/函數(shù)定義 參數(shù)前不需要*符號(hào).下面的例子應(yīng)該很清楚:

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;
}

我無(wú)法理解為什么要對(duì) class 進(jìn)行這種特殊處理.即使是幾乎像一個(gè)類(lèi)的結(jié)構(gòu)也不是這樣使用的.對(duì)象名稱(chēng)是否和數(shù)組一樣被視為地址?

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?

推薦答案

讓您感到困惑的是，聲明為傳遞引用的函數(shù)(使用 &) 不使用實(shí)際地址調(diào)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.

簡(jiǎn)單的答案是將函數(shù)聲明為傳遞引用:

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

void foo(int& x);

就是我們所需要的.然后它會(huì)自動(dòng)通過(guò)引用傳遞.

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

你現(xiàn)在像這樣調(diào)用這個(gè)函數(shù):

You now call this function like so:

int y = 5;
foo(y);

和 y 將通過(guò)引用傳遞.

and y will be passed by reference.

你也可以這樣做(但為什么要這樣做?咒語(yǔ)是:盡可能使用引用，需要時(shí)使用指針):

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;
}

輸出:

yes, &a is b

這篇關(guān)于在 C++ 中通過(guò)引用傳遞對(duì)象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)！