问题
I've searched but couldn't find any results (my terminology may be off) so forgive me if this has been asked before.
I was wondering if there is an easy way to call a void* as a function in C without first declaring a function pointer and then assigning the function pointer the address;
ie. assuming the function to be called is type void(void)
void *ptr;
ptr = <some address>;
((void*())ptr)(); /* call ptr as function here */
with the above code, I get error C2066: cast to function type is illegal in VC2008
If this is possible, how would the syntax differ for functions with return types and multiple parameters?
回答1:
Your cast should be:
((void (*)(void)) ptr)();
In general, this can be made simpler by creating a typedef for the function pointer type:
typedef void (*func_type)(void);
((func_type) ptr)();
I should, however, point out that casting an ordinary pointer (pointer to object) to or from a function pointer is not strictly legal in standard C (although it is a common extension).
回答2:
I get awfully confused when casting to function types. It's easier and more readable to typedef the function pointer type:
void *ptr = ...;
typedef void (*void_f)(void);
((void_f)ptr)();
回答3:
In C++: reinterpret_cast< void(*)() > (ptr) ()
The use of reinterpret_cast saves you a set of confusing parentheses, and the < > clearly sets the type apart from the call itself.
来源：https://stackoverflow.com/questions/2635715/call-a-void-as-a-function-without-declaring-a-function-pointer