可以在函数定义中使用函数原型 typedef 吗? [英] Can a function prototype typedef be used in function definitions?

问题描述

我有一系列具有相同原型的函数，比如说

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

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

现在，我想简化它们的定义和声明.我当然可以使用这样的宏:

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

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

但我想将其保留在 C 中，因此我尝试为此使用存储说明符 typedef:

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

typedef int SpFunc(int a, int b);

这似乎适用于声明:

SpFunc func1; // compiles

但不是定义:

SpFunc func1 {
  // ...
}

这给了我以下错误:

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

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?

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?

注意，gcc 理解我想要做什么，因为，如果我写

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

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

它告诉我

error: function 'func1' is initialized like a variable

这意味着 gcc 理解 SpFunc 是一个函数类型.

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

推荐答案

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:

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:

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.

The identifier declared in a function definition (which is the name of the function) shall have a function type, as specified by the declarator portion of the function definition.

目的是不能从 typedef 继承函数定义中的类型类别:

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

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