我怎样才能的typedef函数指针把它自己类型的函数作为参数？ [英] How can I typedef a function pointer that takes a function of its own type as an argument?

问题描述

示例：取一个函数的函数（带有一个功能（即...）和一个int）和一个int

Example: A function that takes a function (that takes a function (that ...) and an int) and an int.

typedef void(*Func)(void (*)(void (*)(...), int), int);

它爆炸递归其中（...）。有没有这个不能做，或有另一种语法的根本原因？在我看来，它应该有可能不进行强制转换。我真的想通过一个调度表，但是我能明白这一点，如果我可以只通过这一种类型。

It explodes recursively where (...). Is there a fundamental reason this can't be done or is there another syntax? It seems to me it should be possible without a cast. I'm really trying to pass a dispatch-table but I could figure that out if I could just pass this one type.

推荐答案

你可以用函数指针的结构：

You can wrap the function pointer in a struct:

struct fnptr_struct;
typedef void (*fnptr)(struct fnptr_struct *);
struct fnptr_struct {
  fnptr fp;
};

我不知道这是否是在铸造的改进。我怀疑它没有结构是不可能的，因为C需要在使用前必须定义类型和有针对的typedef没有不透明的语法。

I'm not sure if this is an improvement on casting. I suspect that it's impossible without the struct because C requires types to be defined before they are used and there's no opaque syntax for typedef.

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