pointer to 2d array

Question

I have the following code:

int arr[2][2][2]={10,3,4,5,6,7,8,9};
int *p;
printf("%u",arr);
p=(int *)arr;
printf("%u",p);

Which outputs

64166
64164

But I would think that p and arr point to the same memory address. Why are different addresses shown?

I ran the the same piece of code, the result is : 3216608828,3216608828 ! — Rsh
– Rsh, Commented Jul 27, 2012 at 13:09

Jeyaram · Accepted Answer · 2012-07-27 13:06:14Z

2

But same code

 #include <stdio.h>

    int main()
    {

         int arr[2][2][2]={10,3,4,5,6,7,8,9};
         int *p;
         printf("\n%u",arr);
         p=(int *)arr;
         printf("\n%u\n",p);
         return 0;
    }

gives same result only.

enter image description here

answered Jul 27, 2012 at 13:06

Jeyaram

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Comments

Brian · Accepted Answer · 2012-07-27 12:28:06Z

2

Let's walk through the code

 int *p;
 printf("%u",p);

p is an unitialized int pointer. It is going to print out whatever is in memory.

 p=(int *)arr;
 printf("%u",p);

p now is pointing to the address of the array in memory, and prints that address.

answered Jul 27, 2012 at 12:28

Brian

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4 Comments

ghostrider Over a year ago

oh sorry,i made a mistake in the code... the first printf is printf("%u",arr) and since arr and p point to same value then why does the second printf show diff values...

Brian Over a year ago

Are you trying to print out values from the array?

Bo Persson Over a year ago

Adding to this, %u is for printing an unsigned int, %p prints a pointer.

ghostrider Over a year ago

I m just trying to figure out the logic behind this.i dont want to print out values from array.

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