Help with simple pointer program

leroyjenkens

Quote by Mark44

No, **y is not the right expression in scanf, as you know.

As you also know, the value in p is the address you need. What other expression that involves y also contains this address? It's not a very complicated expression.

Is it &*y?

I did a printf("%d", &b)
And a printf("%d", &*y)
And I got back the same answer.

Mark44

You don't need the & operator - *y already is an address.

leroyjenkens

I think I got it. Does this look right?

Code:

#include <stdio.h>
#include <stdlib.h>

int main()
{
   int c;
   int b;
   int a;
   int* x = &a;
   int* p = &b;
   int** y = p;
   int* q = &c;
   int** r = &q;
   int*** z = &r;


    printf("Enter a value for A: ");
    scanf("%d", x);
    printf("Enter a value for B: ");
    scanf("%d", *y);

    *z = *x * **y;
    printf("A * B = %d\n", *z);
    printf("C = %d\nA = %d\nB = %d", *z, *x, **y);

    return 0;
}

Mark44

The problem requirements are satisfied now. Does it give the right results?

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Mark44 Mentor	You don't need the & operator - *y already is an address.