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

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

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

7,1629 gold badges49 silver badges87 bronze badges

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.

Collectives™ on Stack Overflow

pointer to 2d array

2 Answers 2

Comments

4 Comments

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

4 Comments

Related