C语言 可以在函数定义中使用函数原型 typedef 吗？

Can a function prototype typedef be used in function definitions?

提问by bitmask

I have a series of functions with the same prototype, say

我有一系列具有相同原型的函数，比如

int func1(int a, int b) {
  // ...
}
int func2(int a, int b) {
  // ...
}
// ...

Now, I want to simplify their definition and declaration. Of course I could use a macro like that:

现在，我想简化它们的定义和声明。当然，我可以使用这样的宏：

#define SP_FUNC(name) int name(int a, int b)

But I'd like to keep it in C, so I tried to use the storage specifier typedeffor this:

但是我想将它保留在 C 中，因此我尝试为此使用存储说明符typedef：

typedef int SpFunc(int a, int b);

This seems to work fine for the declaration:

这似乎适用于声明：

SpFunc func1; // compiles

but not for the definition:

但不是定义：

SpFunc func1 {
  // ...
}

which gives me the following error:

这给了我以下错误：

error: expected '=', ',', ';', 'asm' or '__attribute__' before '{' token

Is there a way to do this correctly or is it impossible? To my understanding of C this should work, but it doesn't. Why?

有没有办法正确地做到这一点还是不可能？根据我对 C 的理解，这应该有效，但它没有。为什么？

---

---

Note, gcc understands what I am trying to do, because, if I write

注意，gcc 理解我想要做什么，因为，如果我写

SpFunc func1 = { /* ... */ }

it tells me

它告诉我

error: function 'func1' is initialized like a variable

Which means that gcc understands that SpFunc is a function type.

这意味着 gcc 理解 SpFunc 是一个函数类型。

回答by Johannes Schaub - litb

You cannot define a function using a typedef for a function type. It's explicitly forbidden - refer to 6.9.1/2 and the associated footnote:

不能使用函数类型的 typedef 定义函数。这是明确禁止的 - 请参阅 6.9.1/2 和相关的脚注：

The identi?er declared in a function de?nition (which is the name of the function) shall have a function type, as speci?ed by the declarator portion of the function definition.

The intent is that the type category in a function definition cannot be inherited from a typedef:

typedef int F(void); // type F is "function with no parameters
                     // returning int"
F f, g; // f and g both have type compatible with F
F f { /* ... */ } // WRONG: syntax/constraint error
F g() { /* ... */ } // WRONG: declares that g returns a function
int f(void) { /* ... */ } // RIGHT: f has type compatible with F
int g() { /* ... */ } // RIGHT: g has type compatible with F
F *e(void) { /* ... */ } // e returns a pointer to a function
F *((e))(void) { /* ... */ } // same: parentheses irrelevant
int (*fp)(void); // fp points to a function that has type F
F *Fp; //Fp points to a function that has type F

在函数定义中声明的标识符（即函数的名称）应具有函数类型，如函数定义的声明符部分所指定。

目的是函数定义中的类型类别不能从 typedef 继承：

typedef int F(void); // type F is "function with no parameters
                     // returning int"
F f, g; // f and g both have type compatible with F
F f { /* ... */ } // WRONG: syntax/constraint error
F g() { /* ... */ } // WRONG: declares that g returns a function
int f(void) { /* ... */ } // RIGHT: f has type compatible with F
int g() { /* ... */ } // RIGHT: g has type compatible with F
F *e(void) { /* ... */ } // e returns a pointer to a function
F *((e))(void) { /* ... */ } // same: parentheses irrelevant
int (*fp)(void); // fp points to a function that has type F
F *Fp; //Fp points to a function that has type F

回答by user541686

A typedefdefines a type, not a header (which is source code text). You have to use #define(although I don't recommend it) if you need to factor out the code for the header.

Atypedef定义了类型，而不是标题（这是源代码文本）。#define如果您需要分解标题的代码，则必须使用（尽管我不推荐它）。

([Edited] The reason the first one works is that it's not defining a prototype -- it's defining a variable of the type defined by the typedef, which isn't what you want.)

（[编辑] 第一个工作的原因是它没有定义原型——它定义了一个由 定义的类型的变量typedef，这不是你想要的。）