typedef

定义函数指针类型

之前，一直对typedef int(*p)()这种写法定义了p这一类型不清楚。原来是任何声明变量的语句前面加上typedef之后，原来是变量的都变成一种类型。下面是这一解释的原文，很值得一观。

Typedef does not work like typedef [type] [new name]. The [new name] part does not always come at the end.

You should look at it this way: if [some declaration] declares a variable, typedef [same declaration] would define a type.

E.g.:

• int x; declares a variable named x of type int -> typedef int x; defines a type x as int.
• struct { char c; } s; defines a variable named s of some struct type -> typedef struct { char c; } s; defines type s to be some struct type.
• int *p; declares a variable named p of type pointer to int -> typedef int *p; defines a type p as pointer to int.

And also:

• int A[]; declares an array of ints named A -> typedef int A[]; declares a type A as an array of ints.
• int f(); declares a function named f -> typedef int f(); declares a function type f as returning an int and taking no arguments.
• int g(int); declares a function name g -> typedef int g(int); declares a function type g as returning an int and taking one int.

As an aside: Note that all function arguments come after the new name! As those types could be complicated as well, there can be a lot of text after the [new name]. Sadly, but true.

But those are not proper function pointers yet, just function types. I'm not sure a function type exists in C or C++, but it is useful as an intermediate step in my explanation. // 作者并不确定是C/C++否有函数类型这一说

To create a real function pointer, we have to add '*' to the name. Which, sadly, has wrong precedence:

• typedef int *pf(); declares a function type pf as return a int*. Oops, that's not what was intended.

So use () to group:

• typedef int (*pf)(); declares a function pointer type pf as returning an int and taking no arguments.

• typedef int (&rf)(); declares a function reference type rf as returning an int and taking no arguments.

  References:

• C++ Syntax/Semantics Question: Reference to Function and typedef keyword