What is the output of the following code snippet please explain in brief??

Question

for(int i=0;i<8;i++) {
if(i==2) {
i++;
continue; }
else if(5*i==20)
break;
else
cout<<5*i; }

I need explanation of this code and answer

Answer 1

manoj marmat
continue skips the rest of the code and starts from the top of the loop.
break - breaks out of the loop or you could say exits the loop.
So in your code it will iterate like this way
When i = 0 execute else statement so 5*0 so 0 is printed
When i = 1 execute else statement so 5*1 = 5 is printed

Now when value of i = 2 then in the if block value of i is first incremented which is an post increment so the further use of i will take i value to 3 and in the if block continue statement is there which again restart the loop fron top then due to loop increment value of i = 4 so when it goes in else if block value of i will be 5*4 = 20 which make else if condition true and due to use of break statement it break down the loop interation thus the output is 05

Answer 2

Hi. You need to figure out which "if" triggers for every possible "i".
When i=0
1st if - false
2nd if - false
so triggers "else":
cout<<5*i;
When i=1
The same as before:
cout<<5*i;
When i=2
1st if - true
so it triggers:
i++;
continue;
When i=4 (not 3 because of i++ in if and i++ in loop)
1st if - false
2nd if - true
so 2nd if triggers:
break;

Answer 3

I hope you know how a for loop works-
So there are 2 conditions.
1) if i==2 // then i++ and continue.
2) if 5*i == 20 // it's possible when i=4. And break out of the loop.
3) and for rest of the 'i' values print 5*i.

Loop start-
i=0, i<8 true
// else is executed and 5*0 = 0

i=1, i<8 true
// else is executed and 5*1 = 5

i=2, i<8 true
// if is executed and i++ increments i value by 1, so i is 3 and continue starts the next loop.
Since i is already 3 now in for loop i++ will make i=4.

i=4, i<8 true
// else if is executed since 5*4 = 20. The break statement terminates the loop and program execution stops.

Final output should be-
05