C Programming - Arrays - Discussion

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.

Using the name of array directly always gives the base address of array i.e. address of first element in array. also using &arr will also give the address of first element so both the things are same. if someone wants to distinguish between addresses then he should provide position of the array element e.g. &arr[0] or &arr[5].