#include<stdio.h>
int main()
{
int i=0;
for(; i<=5; i++);
printf("%d", i);
return 0;
}
Options
A. 0, 1, 2, 3, 4, 5
B. 5
C. 1, 2, 3, 4
D. 6
Correct Answer
6
Explanation
Step 1: int i = 0; here variable i is an integer type and initialized to '0'. Step 2: for(; i<=5; i++); variable i=0 is already assigned in previous step. The semi-colon at the end of this for loop tells, "there is no more statement is inside the loop".
Loop 1: here i=0, the condition in for(; 0<=5; i++) loop satisfies and then i is incremented by '1'(one) Loop 2: here i=1, the condition in for(; 1<=5; i++) loop satisfies and then i is incremented by '1'(one) Loop 3: here i=2, the condition in for(; 2<=5; i++) loop satisfies and then i is incremented by '1'(one) Loop 4: here i=3, the condition in for(; 3<=5; i++) loop satisfies and then i is increemented by '1'(one) Loop 5: here i=4, the condition in for(; 4<=5; i++) loop satisfies and then i is incremented by '1'(one) Loop 6: here i=5, the condition in for(; 5<=5; i++) loop satisfies and then i is incremented by '1'(one) Loop 7: here i=6, the condition in for(; 6<=5; i++) loop fails and then i is not incremented.
Step 3: printf("%d", i); here the value of i is 6. Hence the output is '6'.
More questions
1. Bitwise can be used to reverse a sign of a number.