What values are j-i and *j-*i actually subtracting ?? The answer i got was 3 30

Question

main()
{
int arr[]={10,20,30,40,50,60};
int *i,*j;
i=&arr[1];
j=&arr[5];
printf("%d %d",j-i,*j-*i);
}

Answer 1

as we know
array is a collection of similar type data having continues memory allocation.............
int arr[]={10,20,30,40,50,60} ;
So what will happen

there is a continues memory allocation of arr
arr[0]. arr[1]. arr[2] arr[3]
[10]. [20]. [30]. [40].

arr[4]. arr[5]
[50]. [60]

let suppose base address of arr[0] index is 1000
than arr[1] is 1002
becoz integer take two byte....

next part in second comment

Answer 2

Second part
pointer store the address of the
so what will happen
when you do this
i=&arr[1]
j=&arr[5]

Here what happen two memory creates which hold the address of
arr[1]
arr[5]

i. j
[1002] [1010]

if we do j-i the and is 8..
now remember this
size of integer is change according to the compiler you used.
in turboc integer take 2 bytes where in gcc integer take 4byte.

as we dont know the base address start at arr[0] we cant predict the answer is 8,4,3;
its total depending upon base address and integer size
...

now second ans.
i -->>1002
j-->>1010

but
*i. focus on the location of arr[1]
*J focus on the location of arr[5]

I.e 20. and 60..

so when you do this*j-*i= 40;

https://code.sololearn.com/cBPoT5vaJjvP/?ref=app

Answer 3

thanks for the help