Help with pointer issue

Firstly, I'm new to C, so please bear with me.

I'm trying to increment a pointer, the following code works correctly with the
following output:

before number = [0], address = [0xbfec32b0}
after number = [1], address = [0xbfec32b4}

Code:

---

#include <stdio.h>

int main(void) {

    int number[2] = { 0,1 };
    int *p_number = NULL;

    p_number = &number[0];

    printf("before number = [%d], address = [%p}\n",*p_number,p_number);

    p_number++;

    printf("after number = [%d], address = [%p}\n",*p_number,p_number);

    return 0;

}

---

However, when I try to perform the same task using a function, I get the following output:

before number = [0], address = [0xbff372a0}
after number = [0], address = [0xbff372a0}

Code:

---

#include <stdio.h>

void increment(int *p_address);

int main(void) {

    int number[2] = { 0,1 };
    int *p_number = NULL;

    p_number = &number[0];

    printf("before number = [%d], address = [%p}\n",*p_number,p_number);

    increment(p_number);

    printf("after number = [%d], address = [%p}\n",*p_number,p_number);

    return 0;

}


void increment(int *p_address) {

    p_address++;

}

---

Any advice with this would be most welcome, I'm positive that I'm doing something wrong.

Thanks
~P

Pointers are values. The value is a memory address. If you need a value change done inside a function, to carry over outside of the function, you need a pointer to the object containing that value. Not the value itself. Consider:

Code:

---

void foo( int x )
{
    x++;
}

---

If we pass a value here, and we increment it, that is lost when the function returns. The reason is, we're not actually passing a variable, we're passing a value. How? Well, because this is legal:

Code:

---

foo( 5 );

---

That happens with everything ever passed to functions. They're all just values. If you need to change a variable's held value, then you need a pointer to that variable, not its value:

Code:

---

void foo( int *x )
{
    (*x)++;
}

---

The same goes for your pointer. You will need a pointer to that pointer.


Quzah.

Look what exactly happens in the function in C. Parameters are passed by value, so a copy is made

Code:

---

void increment(int *p_address) {
    p_address = PARAMETER_PASSED;
    p_address++;

}

---

so if you do

Code:

---

increment(p_number);

---

you have

Code:

---

{
  p_address = p_number;
  p_address++;
}

---

So p_number doesn't increase, only the local variable p_address which is destroyed after the function ends. You have two options. First is to return p_address so you can assign the value, like

Code:

---

void increment(int *p_address) {
    p_address++;
    return p_address;
}

---

and use

Code:

---

p_number = increment(p_number);

---

or pass by reference, like

Code:

---

void increment(int **p_address) {
    (*p_address)++;
}

---

and don't change anything else.
Dunno if this makes sense to you

Quote:

---

Originally Posted by quzah

The same goes for your pointer. You will need a pointer to that pointer.

---

Eg.

Code:

---

increment(&p_number);
void increment(int **p_address) {
    (*p_address)++;
}

---

So you pass the address of the pointer (which is a pointer to the pointer), then you increment the value at that address by dereferencing it.

Sorted. Many Thanks to all for the advice, I've also seen a pratical example of a pointer to a pointer. Excellent.

Thanks
~P