将&变量传递给函数是什么意思？例如，string& insert（size_t pos1，const string& str）; [英] What does it mean to pass a &variable to a function? E.g., string& insert ( size_t pos1, const string& str );

问题描述

我理解传递指针并返回指针：

char * strcat（char * destination，const char * source）;

您正在将包含地址的变量传递给char;返回相同。

但是使用引用运算符传递东西意味着什么呢？或返回？

string& insert（size_t pos1，const string& str）;

我的意思是，我明白发生了什么，我只是不明白符号。为什么不是这样的符号：

string * insert（size_t pos1，const string * str）; // make up

我认为它与传递/返回类的实例有关，但是什么？此语法是否有效;

char& strcat（char& destination，const char& source）; // make up

（所有的函数声明，除了最后的组成两个，都来自 http://www.cplusplus.com ）

解决方案

简单地说，引用是一个指针，而不告诉你它是一个指针。

如果你写下面的简单C：

  void getpi（float * f）
 {
 * f = 3.14; 
} 
 
 float v; 
 getpi（& v）; 
  

您可以使用引用在C ++中编写以下内容：

  void getpi（float& f）
 {
f = 3.14; 
} 
 
 float v; 
 getpi（v）; 
  

它有额外的优点，你可以从正常的by-value参数移动到by-而不对调用者进行更改。假设你有这个：

  class X; //一个很容易复制的小类
 void doSomething（X x）; 
 

但是一段时间后，类X变得很大，你不想通过它的值不再。
在纯C中，您必须将参数更改为指针参数，并更改所有调用者。
在C ++中你可以这样做：

  void doSomething（X& x）; 
  

您不必更改任何来电者。

I understand passing a pointer, and returning a pointer:

char * strcat ( char * destination, const char * source );

You're passing a variable that contains the address to a char; returning the same.

But what does it mean to pass something using the reference operator? Or to return it?

string& insert ( size_t pos1, const string& str );

I mean, I understand what actually happens, I just don't understand the notation. Why isn't the notation this instead:

string * insert ( size_t pos1, const string * str ); //made up

I presume it has something to do with passing/returning the instance of a class, but what? Is this syntax valid; if not why not and if so what does it mean?

char & strcat ( char & destination, const char & source ); //made up

(all of the function declarations, except the last made-up two, are from http://www.cplusplus.com )

解决方案

Simply said, a reference is a pointer without telling you it's a pointer.

If you would write the following in plain C:

void getpi (float *f)
{
*f = 3.14;
}

float v;
getpi(&v);

You can write the following in C++ using references:

void getpi (float &f)
{
f = 3.14;
}

float v;
getpi(v);

It has the additional advantage that you can move from normal by-value argument to by-reference argument, without making changes to the caller. Suppose you have this:

class X;   // A rather small class that can be easily copied
void doSomething (X x);

But after a while, class X becomes really big and you don't want to pass it by value anymore. In plain C you have to change the argument to a pointer argument, and change all the callers. In C++ you can simply do this:

void doSomething (X &x);

And you don't have to change any of the callers.

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