What will be the output of the following program?
#include <stdio.h>
int main()
{
int i;
for (i = 0; i++<10;)
printf("%d", i++);
return 0;
}

Correct option is C

The program uses the post-increment operator (i++) both in the for loop condition and inside the printf() statement. Initially, i = 0. In the first condition, i++ < 10 compares 0 with 10 and then increments i to 1; therefore, printf("%d", i++) prints 1 and increments i to 2. The same process repeats, causing the values 1, 3, 5, 7 and 9 to be printed. After printing 9, i becomes 10, and the next condition 10 < 10 becomes false. Hence, the output is 13579.
Information Booster
1. Post-increment (i++)
· The post-increment operator first uses the current value of the variable and then increments it by 1.
· For example:
int i = 5;
printf("%d", i++);
prints 5, after which i becomes 6.
2. Evaluation of the for condition
· The condition is:
i++ < 10
· The comparison uses the old value of i.
· After the comparison, i is incremented.
· Thus, every successful condition check increases i by 1 before entering the loop body.
3. Evaluation inside printf()
· The statement:
printf("%d", i++);
prints the current value and then increments i.
· Therefore, the value of i increases twice per successful iteration: once in the condition and once in the loop body.
4. Iteration-wise execution