C Programming - Arrays - Discussion

@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).

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));
}

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

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

Someone explain this.