main() { int arr[]={12,14,16,18,20} printf("%u%u";arr+1,&arr+1

main()
{
int arr[]={12,14,16,18,20}
printf("%u%u";arr+1,&arr+1);
}
//tell me the answer of this question with proper reason..:)//

Question Posted / arth gupta

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main() { int arr[]={12,14,16,18,20} printf("%u%u";arr+1,&arr+1); } //tell me the..

Answer / rpr

The correct question should be:

main()
{
int arr[]={12,14,16,18,20};
printf("%u%u",arr+1,&arr+1);
}

there are two different notations:
arr+1 and &arr+1;

arr is same as &arr[0] - location of arr[0], first element
arr+1 is same as &arr[1] - location of arr[1]
arr+2 is same as &arr[2] - location of arr[2]
and so on...

here if address of arr is 6524 then address of arr+1 is
6526, address of arr+2 is 6528.. as arr is of int type so a
difference of two.

now about &arr
it represents the location of entire array.
so it will gives address of first element of array.
when 1 is added to it, the size of array is added to it..

example..

if the address of first element,arr[0] is 6524
and the size of array is-
-(no. of elements)*(size of each element)
here 5*2=10

then,
&arr = 6524
&arr+1 = 6524+(1*size of array)=6524+(1*10)=6534
&arr+2 = 6524+(2*10)=6544
and so on..

one more point:
as here notations are behaving in a pointer manner(actually
reference).. you should know pointer saves values in
unsigned int- from 0-65535
..
once the value of pointer reaches 65535, it comes back to 0.

-rpr..ravidasonline@gmail.com

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