Question

In “void pointers” example in the tutorial on cplusplus.com, I try to compare like following. Why do we still need * in parenthesis? What is happening when no *?

void increase(void* data, int psize) {
    if (psize == sizeof(char)) {
        char* pchar;
        pchar = (char*) data;
        cout << "pchar=" << pchar << endl;
        cout << "*pchar=" << *pchar << endl;
        //++(*pchar);   // increases the value pointed to, as expected
        ++(pchar);      // the value pointed to doesn't change 
    } else if (psize == sizeof(int)) {
        int* pint;
        pint = (int*) data;
        //++(*pint);   // increases the value pointed to, as expected
        ++(pint);      // the value pointed to doesn't change 
    }
}

int main() {
    char a = 'x';
    int b = 1602;
    increase(&a, sizeof(a));
    increase(&b, sizeof(b));
    cout << a << ", " << b << endl;
    return 0;
}

---

Update after accepting solution) I try to make clear what I didn’t get, based on @Cody Gray’s answer. The address of pchar is incremented to point to nonsense location. But because variable a in main is coutted instead of pchar, this cout still prints a value that somewhat makes sense (in this example it would be ‘x’).

Answer 1

The * operator dereferences the pointer.

Therefore, this code:

++(*pint)

increments the value pointed to by pint.

By contrast, this code:

++(pint)

increments the pointer itself, pint. This is why the value pointed to by the pointer doesn’t change. You’re not changing the value pointed to by the pointer, but rather the value of the pointer itself.

Incrementing a pointer will cause it to point to an entirely different value in memory. In this case, since you only allocated space for a single integer, it will point to a nonsense value.