奇怪类型在c ++ [英] Strange type in c++

问题描述

我有一个方法与原型：

  bool getAssignment（const Query& query，Assignment *& result）; 
  

我有点困惑第二个参数的类型（ Assignment * & result ），因为我不认为我曾经看到过类似的东西。它的用法如下：

 赋值* a; 
 if（！getAssignment（query，a））
 return false; 
  

它是指针的引用还是其他方法？或两者都不？任何解释赞赏。谢谢。

解决方案

这是一个指针的引用。这个想法是能够改变指针。

$ b pre> void f（char * p）
{
p = new char [100];
}
int main（）
{
char * p_main = NULL;
f（p_main）;
return 0;
}


不会更改 p_main 指向分配的字符数组（它是一个明确的内存泄漏）。这是因为你复制指针，它通过值传递（这就像传递一个 int 的值;例如 void f（int x）！= void f（int& x））

所以，如果你改变 f ：

  & p）
  

现在，这将通过 p_main 通过引用并将改变它。因此，这不是内存泄漏，在执行 f 后， p_main 将正确指向分配的内存。

---

PS同样可以通过使用双指针（例如， C 没有引用）来完成：

  void f（char ** p）
 {
 * p = new char [100]; 
} 
 int main（）
 {
 char * p_main = NULL; 
 f（& p_main）; 
 
 return 0; 
} 
  

I have a method with the prototype:

 bool getAssignment(const Query& query, Assignment *&result);

I am a bit confused about the type of the second param (Assignment *&result) since I don't think I have seen something like that before. It is used like:

 Assignment *a;
 if (!getAssignment(query, a))
    return false;

Is it a reference to a pointer or the other way around ? or neither ? Any explanation is appreciated. Thanks.

解决方案

It's a reference to a pointer. The idea is to be able to change the pointer. It's like any other type.

---

Detailed explanation and example:

void f( char* p )
{
    p = new char[ 100 ];
}
int main()
{
    char* p_main = NULL;
    f( p_main );
    return 0;
}

will not change p_main to point to the allocated char array (it's a definite memory leak). This is because you copy the pointer, it's passed by value (it's like passing an int by value; for example void f( int x ) != void f( int& x ) ) .

So, if you change f:

void f( char*& p )

now, this will pass p_main by reference and will change it. Thus, this is not a memory leak and after the execution of f, p_main will correctly point to the allocated memory.

---

P.S. The same can be done, by using double pointer (as, for example, C does not have references):

void f( char** p )
{
    *p = new char[ 100 ];
}
int main()
{
    char* p_main = NULL;
    f( &p_main );

    return 0;
}

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