可能的重复:
c++ * vs &在函数声明中
我知道对于你们中的许多人来说，这可能是一个非常基本的问题，但我真的很难找到一个好的、彻底的解释，尽管我尽了最大的谷歌搜索.我确信答案就在那里，所以我的搜索词一定很糟糕.
I know that this probably seems like an incredibly elementary question to many of you, but I have genuinely had an impossible time finding a good, thorough explanation, despite all my best Googling. I'm certain that the answer is out there, and so my search terms must be terrible.
在 C++ 中，各种符号及其组合用于标记参数(以及这些参数的实参).它们的确切含义是什么?
In C++, a variety of symbols and combinations thereof are used to mark parameters (as well as arguments to those parameters). What, exactly, are their meanings?
例如:void func(int *var) 和 void func(int **var) 有什么区别?int &var 怎么样?
Ex: What is the difference between void func(int *var) and void func(int **var)? What about int &var?
同样的问题代表返回类型，以及参数.int& 有什么作用?func(int var) 和 int* func(int var) 相比是什么意思?在参数方面，y = func(*x) 与 y = func(&x) 有何不同?
The same question stands for return types, as well as arguments. What does int& func(int var) mean, as compared to int* func(int var)? And in arguments, how does y = func(*x) differ from y = func(&x)?
如果你能指出我正确的方向，我很乐意阅读关于这个主题的大量书籍.此外，我对通用编程概念非常熟悉:OO、泛型/模板等，只是不熟悉 C/C++ 中使用的符号.
I am more than happy to read enormous volumes on the subject if only you could point me in the right direction. Also, I'm extremely familiar with general programming concepts: OO, generics/templates, etc., just not the notation used in C/C++.
似乎我给人的印象是我不知道什么是指针.我想知道这怎么可能:)
It seems I may have given the impression that I do not know what pointers are. I wonder how that could be :)
所以澄清一下:我完全理解指针是如何工作的.我没有理解，并且奇怪地无法找到答案的是，例如void func(int &var)"的含义.在赋值语句的情况下，&"运算符将在右侧，如int* x = &y;"，但在上面，&"操作员实际上位于左侧.换句话说，它对 l 值进行操作，而不是对 r 值进行操作.这显然不能具有相同的含义.
So for clarification: I understand perfectly how pointers work. What I am not grasping, and am weirdly unable to find answers to, is the meaning of, for example 'void func(int &var)'. In the case of an assignment statement, the '&' operator would be on the right hand side, as in 'int* x = &y;', but in the above, the '&' operator is effectively on the left hand side. In other words, it is operating on the l-value, rather than the r-value. This clearly cannot have the same meaning.
我希望我现在更有意义了吗?
I hope that I'm making more sense now?
要理解这一点，您首先需要了解指针和引用.假设您已经知道什么是指针和引用，我将简单地解释您所询问的类型声明语法.
To understand this you'll first need to understand pointers and references. I'll simply explain the type declaration syntax you're asking about assuming you already know what pointers and references are.
在 C 中，据说声明跟随使用".这意味着声明变量的语法模仿使用变量:通常在声明中，您将拥有像 int 或 float 这样的基本类型，其后跟一些看起来像表达式的东西.例如，在 int *y 中，基本类型是 int 并且表达式相似是 *y.此后，该表达式的计算结果为具有给定基类型的值.
In C, it is said that 'declaration follows use.' That means the syntax for declaring a variable mimics using the variable: generally in a declaration you'll have a base type like int or float followed something that looks like an expression. For example in int *y the base type is int and the expression look-alike is *y. Thereafter that expression evaluates to a value with the given base type.
所以 int *y 意味着后面的表达式 *y 是一个 int.这意味着 y 必须是一个指向 int 的指针.这同样适用于函数参数，实际上也适用于整个函数声明:
So int *y means that later an expression *y is an int. That implies that y must be a pointer to an int. The same holds true for function parameters, and in fact for whole function declarations:
int *foo(int **bar);
在上面的 int **bar 说 **bar 是一个 int，暗示 *bar 是一个指向 int 的指针，并且bar 是一个指向 int 指针的指针.它还声明 *foo(arg) 将是一个 int(给定适当类型的 arg)，暗示 foo(arg) 结果在指向 int 的指针中. 所以整个函数声明读作foo 是一个函数，它接受一个指向 int 的指针的指针，并返回一个指向 int 的指针."
In the above int **bar says **bar is an int, implying *bar is a pointer to an int, and bar is a pointer to a pointer to an int. It also declares that *foo(arg) will be an int (given arg of the appropriate type), implying that foo(arg) results in a pointer to an int. So the whole function declaration reads "foo is a function taking a pointer to a pointer to an int, and returning a pointer to an int."
C++ 添加了引用的概念，并在此过程中稍微混淆了 C 风格的声明.因为使用地址运算符 & 获取变量的地址必须产生一个指针，所以 C 在声明中对 & 没有任何用处；int &x 表示 &x 是一个 int，暗示 x 是某种类型，其中获取该类型的地址会导致int. 所以因为这个语法没有被使用，C++ 将它用于一个完全不同的目的.
C++ adds the concept of references, and messes C style declarations up a little bit in the process. Because taking the address of a variable using the address-of operator & must result in a pointer, C doesn't have any use for & in declarations; int &x would mean &x is an int, implying that x is some type where taking the address of that type results in an int. So because this syntax is unused, C++ appropriates it for a completely different purpose.
在 C++ 中 int &x 表示 x 是对 int 的引用.使用变量不涉及任何运算符来取消引用"引用，因此引用声明符符号与地址运算符冲突并不重要.同一个符号在两种语境中的含义完全不同，在允许另一种意义的语境中，永远不需要使用一种意义.
In C++ int &x means that x is a reference to an int. Using the variable does not involve any operator to 'dereference' the reference, so it doesn't matter that the reference declarator symbol clashes with the address-of operator. The same symbol means completely different things in the two contexts, and there is never a need to use one meaning in the context where the other is allowed.
所以 char &foo(int &a) 声明了一个函数，它接受对 int 的引用并返回对 char 的引用.func(&x) 是一个将 x 的地址传递给 func 的表达式.
So char &foo(int &a) declares a function taking a reference to an int and returning a reference to a char. func(&x) is an expression taking the address of x and passing it to func.
1.事实上，在用于声明函数的原始 C 语法中，更严格地遵循声明跟随使用".例如，你将一个函数声明为 int foo(a,b) 并且参数的类型在别处声明，这样声明看起来就像一个 use 一样，没有额外的类型名.子>
1. In fact in the original C syntax for declaring functions 'declarations follow use' was even more strictly followed. For example you'd declare a function as int foo(a,b) and the types of parameters were declared elsewhere, so that the declaration would look exactly like a use, without the extra typenames.
2.当然 int *&x; 可能有意义，因为 *&x 可能是一个 int，但 C 实际上并没有这样做.
2. Of course int *&x; could make sense in that *&x could be an int, but C doesn't actually do that.
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