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        函数声明中的“*&amp;"是什么意思?

        时间:2023-12-03
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                  本文介绍了函数声明中的“*&amp;"是什么意思?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

                  问题描述

                  我写了一个类似这样的函数:

                  void myFunc(myStruct *&out) {out = 新的 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(数据);

                  这将填充data中的所有字段.如果我在声明中省略了 '&',这将不起作用.(或者更确切地说,它只会在函数中本地工作,但不会改变调用者的任何内容)

                  有人可以向我解释这个*&"实际上是做什么的?看起来很奇怪,我就是无法理解.

                  解决方案

                  The &C++ 变量声明中的符号表示它是一个引用.>

                  它恰好是一个指针的引用,它解释了你所看到的语义;被调用的函数可以改变调用上下文中的指针,因为它有一个对它的引用.

                  所以,重申一下,这里的操作符号"不是 *&,这种组合本身并不意味着很多.*myStruct * 类型的一部分,即指向 myStruct 的指针",而 & 使得它是一个引用,因此您可以将其读作out 是对指向 myStruct 的指针的引用".

                  在我看来,最初的程序员可以这样写:

                  void myFunc(myStruct * &out)

                  甚至(不是我个人的风格,但当然仍然有效):

                  void myFunc(myStruct* &out)

                  当然,关于风格还有很多其他的意见.:)

                  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. :)

                  这篇关于函数声明中的“*&amp;"是什么意思?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持跟版网!

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