Simple Question: Pointers and Uninitialized Variables Explained

[yungman]

Why this doesn't work?

#include<iostream>
using namespace std;
int main()
{
    int* p;
    *p = 1;//error said uninitialized variable p used.
    count << *p << endl;
    return 0;
}

I know if I do this, it works:

#include <iostream>
using namespace std;

int main ()
{
    int* p;
    int x = 10;
    p = &x;
    count << *p << endl;

  return 0;
}

I thought when you declare a pointer int*p; you already allocate a memory for an integer pointed by p already.

Or int*p; only allocate memory for pointer p to store the address, that the address is not valid until using p = &x; to write the address of x into p?
Thanks

 

[jtbell]

Or int*p; only allocate memory for pointer p to store the address, that the address is not valid until using p = &x; to write the address of x into p?

Correct. Compare this to writing simply int x;. This says simply that x is a variable that is intended to contain an int. It doesn't store any particular value into x. At this point, x contains, in effect, a random bit pattern.

When you write simply int *p; this says that p is a variable that is intended to contain a pointer to an int. It doesn't store a particular value into p, that is, it doesn't make p point to any memory location in particular. At this point, p contains, in effect, a random bit pattern.

 

[yungman]

Thanks Jtbell, that clarified for me.

 

[jtbell]

the address is not valid until using p = &x; to write the address of x into p?

That's one way to put a valid address into p. Another way, of course, is to allocate "new" memory explicitly. p = new int; allocates memory intended to hold an int and stores its address in p, but doesn't store any particular value in the newly-allocated memory, so in effect, *p gives you a random bit pattern.

To put something into the newly-allocated memory, you can initialize it when you allocate it: int *p = new int(x); declares p to be a pointer intended to point to an int, allocates an int-sized chunk of memory, and finally copies the value of x into the newly-allocated memory. Or you can initialize the new memory later: int *p = new int; *p = x;. Or you can separate all three steps: int *p; p = new int; *p = x;.