Question

QUESTION: Explain the output of following C code by showing dry run for each step.

main( )
{
int a[5] = { 5, 1, 15, 20, 25 } ;
int i, j, k = 1, m ;
i = ++a[1] ;
j = a[1]++ ;
m = a[i++] ;
printf ( "\n%d %d %d", i, j, m ) ;
}

Answer 1

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Each statement is explained in comments

#include<stdio.h>
#include<stdlib.h>
main( )
{
// array declaration where index starts from 0 to 4
// a[0]=5, a[1]=1, a[2]=15, a[3]=20, a[4]=25
int a[5] = { 5, 1, 15, 20, 25 } ;
// declaration of variables
int i, j, k = 1, m ;
// here a[1] is pre-increment so first a[1] is incremented and then stored in i
// so a[1] is 2 and i = 2
i = ++a[1] ;
// here a[1] post-increment so first value is stored in j then increment in value of a[1]
// so a[1] =2 before increment so value of j is stored first i.e. 2 then increment in a[1] so a[1] = 3 now
j = a[1]++ ;
// here i is post-incremented first this statement is incremented then value so i is incremented
// so i before statement is 2 and a[2] is 15 so m = 15 and i after statement is incremented so i =3 now
m = a[i++] ;
// i = 3 and j=2 and m = 15 will be printed
printf ( "\n%d %d %d", i, j, m ) ;
return 0;
}

Screenshot of output :-