指向函数返回函数指针的指针

I would like to declare a variable of type pointer to function returning pointer to function. Essentially what the following does, but without any typedefs:

我想声明一个类型指针的变量指向函数返回指向函数的指针。基本上是以下内容,但没有任何typedef:

typedef void (*func)();
typedef func (*funky_func)();

funky_func ptr;

I tried the following

我尝试了以下内容

(void (*)()) (*ptr)();

but it gives an "undeclared identifier"-error for ptr (probably due to completely different parsing). Being not that well-versed in the intricacies of parsing C++, I'd like to know if this is even possible and if yes, how to do it.

但它为ptr提供了一个“未声明的标识符” - 错误(可能是由于完全不同的解析)。由于不熟悉解析C ++的复杂性,我想知道这是否可行,如果是,那该怎么做。

(Please consider this an entirely artificial scenario for the sake of curiosity, without any practical reason. I am perfectly aware that in practice typedefs are the way to go here, if using function pointers at all.)

(为了好奇,请考虑这是一个完全人为的场景,没有任何实际原因。我完全知道在实践中,如果使用函数指针,那么typedef就是这里的方法。)

2 个解决方案

#1

The general rule of C (and C++) declarations is: if you type the declaration as an expression, it will have the declaration's type.

C(和C ++)声明的一般规则是:如果您将声明类型化为表达式,它将具有声明的类型。

So, you want a pointer to function which returns pointer to function returning void.

所以,你想要一个指向函数的指针,它返回返回void的函数的指针。

Let's say we have such a pointer, ptr. How to get void out of it?

假设我们有这样一个指针,ptr。如何摆脱它?

Dereference ptr, getting a function returning pointer to function returning void: *ptr

取消引用ptr,获取函数返回指向函数的指针返回void:* ptr
Call the function, getting a pointer to function returning void: (*ptr)()

调用函数,获取函数返回void的指针:(* ptr)()
Dereference that pointer, getting a function returning void: *(*ptr)()

取消引用指针,得到函数返回void:*(* ptr)()
Call that function, getting void: (*(*ptr)())()

调用该函数,得到无效:(*(* ptr)())()

Now just turn this into a declaration:

现在把它变成一个声明:

void (*(*ptr)())();

P.S. I know other have answered in the meantime (and I've upvoted). But I wanted to show the generic process to arrive at the declaration form.

附:我知道其他人在此期间已经回答了(我已经投了赞成票)。但我想展示通用流程以获得申报表。

#2

You can have a look at the declaration of signal() which is a function taking a void(*)() and returning one of those. The variable ptr can be declared like this:

你可以看一下signal()的声明,它是一个带有void(*)()并返回其中一个的函数。变量ptr可以这样声明:

void (*(*ptr)())()

The notation is a bit awkward and clearly inside out. It may be easier to use trailing return types:

这种符号有点尴尬,显而易见。使用尾随返回类型可能更容易:

auto (*ptr)() -> void (*)()

... or, of course, use trailing return types all the way through:

...或者,当然,一直使用尾随返回类型:

auto (*ptr)() -> auto (*)() -> void

#1