未知类型的C ++算术：无效指针？

Question

(I want to deal with "void" I am avoid "template class" because it is too much trouble)

ok establish some ground

int aa=3;
int bb=4;
int cc=0;

int *a, *b, *c;

a = &aa;
b = &bb;
c = &cc;

*c = *a +*b; //yields 7

but I want to do the following:

void *va;
void *vb;
void *vc;

va = a;
vb = b;

*c = (int)(*va + *vb);  // <-- 3 errors see below

but I get errors:

 error C2100: illegal indirection
 error C2100: illegal indirection
 error C2110: cannot add two pointers

Answer 1

You can't dereference void pointer.

If you want to turn them into int* you should do that before dereferencing;

(*(int*)vc) = (*(int*)va) + (*(int*)vb);

But you better be more specific, what exactly you want to do that you call "the following".

Answer 2

You cannot do many operations at all on truly unknown types in C or C++.

You could use Boost Variant to support multiple arithmetic types in one variable, or you could use a union and basically do what Boost Variant does by discriminating it yourself. Or you could build a type hierarchy and have virtual methods to perform the operations you need. But you can't perform arithmetic on values whose types have been completely erased .

Answer 3

The void pointer is a special pointer, it specifically denotes the absence of a type. These are often used to provide storage to a caller who does know the type and can cast back and forth.

Dereferencing a void* makes no sense and is, in-fact, invalid syntax, the whole point of dereferencing a pointer is to get the item being pointed to, if you don't specify what you're pointing at, you cant get it.

To "fix" your code you'd need to cast before dereferencing: http://www.ideone.com/eKxxn

#include <iostream>

int main() {
        int ia = 1, ib = 2, ic = -1;

        int *a = &ia, *b = &ib, *c = &ic;
        void *va, *vb;

        va = a;
        vb = b;

        *c = (int)(*static_cast<int*>(va) + *static_cast<int*>(vb));
        std::cout << *c; // Outputs "3"
}