C Programming - Arrays - Discussion

Please explain me clearly to get the answer.

Someone explain this.

When I'm running an example of this.. &arr and arr are giving the same value. How?

Can anyone explain it in depth. not able to get what they r saying..

Try this

#include<stdio.h>
int main()
{
int arr[2];
printf("%d \n",arr);
printf("%d \n",(arr+1));
printf("%d \n",&arr);
printf("%d \n",(&arr+1));
}

@Ram:You are right ,but the one thing here need to be consider is that, it also depends upon the dimensions of an array.you gave example of 1 D array.But problem is for @ D array.

In 2D array,an array in argument acts as a pointer(compiler dependent).so a(a+0) points to 1st row and a+1 points to second row and it's first element.

And &arr gives the address of array of ints(address of last element of an array which is equivalent to base address+size of type * number of elements).

Both value will be same only the scaling factor will change.

So out put will be same and the answer will be A as yes.

arr and &arr are same.
both returns the base address of the arr at any time.
for better understanding execute ram's program.
but for 2D array arr,&arr[0] both returns the same address.

They are same both arr and &arr holding starting address of array.

Considering an Array &arr and arr returns the address of the first element of the array and the result goes same. And the answer is "YES" I believe.