C如何从va_arg获取类型为函数的指针的参数

Question

I know that if I passed an argument like void (*func)(void *) to a variadic function, I can retrieve the argument like:

void (*func)(void *) = va_arg( args, void (*)(void) );

What if I pass something like void (** func)(void *) ? What is the correct syntax to retrieve an argument of this type using va_arg ?

Answer 1

Being frankly, your code is not standard-compliant. There is a tiny restriction for second argument of va_arg() macro:

... The parameter type shall be a type name specified such that the type of a pointer to an object that has the specified type can be obtained simply by postfixing a * to type. ...

According to this, notation like void (*)(void *) is unacceptable in this case. Since simple appending of * won't give you pointer to pointer to function. You may use only typedef -ed aliases:

typedef void (*func_ptr)(void *);
typedef void (**ptr_to_func_ptr)(void *);

func_ptr var1 = va_arg(ap, func_ptr);
ptr_to_func_ptr var2 = va_arg(ap, ptr_to_func_ptr);

Answer 2

Same as you've mentioned:

typedef void (** func_t)(void *);
func_t ppf;
va_list vl;
va_start(vl,n);
ppf = va_arg(vl, func_t);
...

Answer 3

To help pointer-to-function, I always use typedef as follows:

typedef void VoidFn(void *); // Function accepts void * and returns nothing.

That declares the prototype of the function as a typedef , allowing the use of the typedef elsewhere.

That way if you have this function:

void SomeFn(void *) {...}

you can declare this pointer:

VoidFn *fnPtr = &SomeFn; // A pointer to such a function.

This then makes it easier to change the prototype independently of the pointer, helping more... sophisticated constructs:

typedef void *VoidPtrFn(void *); // Function takes a void *, returning a void *

void *SomeOtherFn(void *) { ... }
VoidPtrFn *otherFnPtr = &SomeOtherFn;
VoidPtrFn **otherFnPtrPtr = &otherFnPtr;