How to typecast a void pointer to an int pointer and then store an int in it?

Question

1.    void* x;
2.    x = new int [10];
3.    x = static_cast<int*>(x);
4.    *x = 2;

On line 4, I am getting: error: 'void*' is not a pointer-to-object type

Answer 1

You need to define the new pointer type.

You static casted x, but the type information is lost after the static cast since at declaration time x is void*.

x will persist to be void* throughout his lifetime.

Here's a working code sample :

     void* x;
     int* ptr;
     x = new int[10];
     ptr = static_cast<int*>(x);
     *ptr = 2;

OR alternativelly, you can asign and cast in the same line :

    *(static_cast<int*>(x)) = 2;

Answer 2

Since any pointer (except pointer to member and pointer to functions, which are something else entirely) can be implicitly converted to a void pointer, your line 2 implicitly converts the result of new int[10] from int * to void * .

Similarly, your line 3 explicitly converts x to int * (the static_cast ) and the result of that is implicitly converted back and stored in x . That has no net effect. If the compiler is smart enough, it would ignore that statement completely.

What you need to do is introduce another variable which is a pointer to int .

void *x;
x = new int[10];
int *y = static_cast<int *>(x);
*y = 2;

If you really want to use a void pointer without any variable that is an int pointer, do this;

void *x;
x = new int[10];
*(static_cast<int *>(x)) = 2;

That is exceedingly ugly, and only needed in specialised circumstances.

In practice, unless you need the void pointer for something else in particular, eliminate it completely.

int *x;
x = new int[10];
*x = 2;

The fact that there is no need to monkey around with any explicit type conversions makes this less error prone too.