如何定义一个函数指针，该函数指针具有与给定函数相同的参数并返回值？

Question

For example, I have a function foo :

int foo(int a, int b)
{
    return a + b;
}

I can define a function pointer:

int (*pfoo)(int, int);

But how can I do this dynamically in program?
I want a function that takes a function as a parameter, and return a function pointer that takes the same arguments and return value as a given function.
Then I can use it like this:

void* pfoo = getFuncPtrFromFunc(foo);  

Which does what the code above did.

Is this possible?

Answer 1

You cannot do this at run-time (ie dynamically); remember that all types (and function pointers are types) are statically determined at compile-time.

Imagine if you could do this, and you somehow obtained a function-pointer whose type was variable (ie it could point to a float (*)(int,int) or a char (*)(float) ). How would you ever be able to call that and provide a meaningful list of arguments?

You can, however, get this information at compile-time ; one way is to use the Boost TypeTraits library .

Answer 2

c ++是一种静态类型的语言，不允许您做自己想做的事情。