Casting from void* to an object array in c++

Question

I'm having problems getting this to work,

class A {
public:
    A(int n) {
        a = n;
    }
    int getA() {
        return a;
    }
private:
    int a;
};

int main(){

    A* a[3];
    A* b[3];

    for (int i = 0; i < 3; ++i) {
        a[i] = new A(i + 1);
    }

    void * pointer = a;

    b = (A* [])pointer;  // DOESNT WORK Apparently ISO C++ forbids casting to an array type ‘A* []’.
    b = static_cast<A*[]>(pointer); // DOESN'T WORK invalid static_cast from type ‘void*’ to type ‘A* []’

    return 0;
}

And i can't use generic types for what i need.

Thanks in advance.

The variable b is an array of 3 pointers. Why do you want a cast a void pointer into it? — Arun
– Arun, Commented Oct 12, 2010 at 17:22

sbi · Accepted Answer · 2011-01-20 23:57:13Z

10

Arrays are second-class citizen in C (and thus in C++). For example, you can't assign them. And it's hard to pass them to a function without them degrading to a pointer to their first element.
A pointer to an array's first element can for most purposes be used like the array - except you cannot use it to get the array's size.

When you write

void * pointer = a;

a is implicitly converted to a pointer to its first element, and that is then casted to void*.

From that, you cannot have the array back, but you can get the pointer to the first element:

A* b = static_cast<A*>(pointer);

(Note: casting between pointers to unrelated types requires a reinterpret_cast, except for casts to void* which are implicit and from void* to any other pointer, which can be done using a static_cast.)

edited Jan 20, 2011 at 23:57

answered Oct 12, 2010 at 17:14

sbi

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Comments

Ben Voigt · Accepted Answer · 2010-10-12 17:36:00Z

9

Perhaps you mean to do

memcpy(b, (A**)pointer, sizeof b);

?

A static_cast version is also possible.

answered Oct 12, 2010 at 17:36

Ben Voigt

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1 Comment

Armen Tsirunyan Over a year ago

@Carlos: If you find this answer was what you needed, please vote it up and accept it (the check under the votes). Thank you.

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