explain this code

Question

#include <stdio.h>
int main()
{
int arr[]={2,1,9,20};
int a,b,c;
a=++arr[1];
b=arr[1]++;
c=arr[a++];
printf("%d %d %d",a,b,c);
return 0;
}

Answer 1

Arb Rahim Badsa

You are right but accidentally typed arr[3] instead of arr[2] for number 9, oops!!

a++ is 2 because post increment, no?

Answer 2

Here is the explanation :))

a =++arr[1];

arr[1] is 1 and ++1 is 2. So,
a = 2; However, arr[1] is now 2 (not 1 anymore) as well.

b = arr[1]++;

arr[1] is now 2 and we have used post increment, so it will be as it is (But it will be increased further). Means that,

b = 2;

Now,

c = arr[a++];

a is 2. Because we have used post increment
(but will be increased further).

Therefore, arr[a++] is arr[2]. Which is 9. At last we have :

a = 3;
b = 2;
c = 9;

Therefore the outputs are 3,2 & 9 :))

Answer 3

Gen2oo Really thanks bro :))

Answer 4

Arb Rahim Badsa

Your explanation is good +1.

Answer 5

Gen2oo
Thanks everybody.
But why b is not print 3?