函数声明中的`*&`是什么意思？ [英] What does `*&` in a function declaration mean?

问题描述

我写了一个沿着这一行的函数：

  void myFunc（myStruct *& out）{
 out = new myStruct; 
 out-> field1 = 1; 
 out-> field2 = 2; 
} 
  

现在在调用函数中，我可能会这样写：

  myStruct * data; 
 myFunc（data）; 
  

将填充 data 。如果我在声明中省略& ，这将不起作用。 （或者，它只在本地函数中工作，但不会更改调用方中的任何内容）

有人可以向我解释这个 *& '实际上是什么？

解决方案

& C ++变量声明中的符号表示它是参考。

它恰好是一个指针的引用，它解释了你看到的语义;被调用函数可以改变调用上下文中的指针，因为它具有对它的引用。

因此，为了重申，这里的操作符号不是 *& ，这种组合本身并不意味着很多。 * 是类型 myStruct * 的一部分，即指向 myStruct ，并且& 使它成为引用，所以你读它为 out 是对 myStruct 的指针的引用。

原来的程序员在我看来可以帮助写为：

  void myFunc（myStruct *& out）
  pre> 
 
 
甚至（不是我的个人风格，但当然仍然有效）：
 code> void myFunc（myStruct *& out）
  
当然，关于风格的意见。 ：）
 
I wrote a function along the lines of this:
void myFunc(myStruct *&out) {
    out = new myStruct;
    out->field1 = 1;
    out->field2 = 2;
}
Now in a calling function, I might write something like this:
myStruct *data;
myFunc(data);
which will fill all the fields in data. If I omit the '&' in the declaration, this will not work. (Or rather, it will work only locally in the function but won't change anything in the caller)

Could someone explain to me what this '*&' actually does? It looks weird and I just can't make much sense of it.
解决方案
The & symbol in a C++ variable declaration means it's a reference.

It happens to be a reference to a pointer, which explains the semantics you're seeing; the called function can change the pointer in the calling context, since it has a reference to it.

So, to reiterate, the "operative symbol" here is not *&, that combination in itself doesn't mean a whole lot. The * is part of the type myStruct *, i.e. "pointer to myStruct", and the & makes it a reference, so you'd read it as "out is a reference to a pointer to myStruct".

The original programmer could have helped,  in my opinion, by writing it as:
void myFunc(myStruct * &out)
or even (not my personal style, but of course still valid):
void myFunc(myStruct* &out)
Of course, there are many other opinions about style. :)

                        
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