Clash Royale CLAN TAG

Can't understand output from this C program [on hold]

Here is the question that I attempted in a mock test.

main(){

int a;

a=1;

while(a<=10){
printf("%d",a);
if(a>3 && a<8)
continue;
a++;
}
printf("%d",a+10);
}


Options are as follows

(A) 1 2 3 4 5 6 ...................infinite
(B) 1 2 3 4 5 5 ...................infinite
(C) 1 2 3 4 4 4 ...................infinite
(D) 1 2 3 4 4 3 ...................infinite
Answer : 1 2 3 4 4 4 ...................infinite


I did not get how the answer is (C). Can anyone explain?

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continue

a

a

a

a>3 && a<8

continue

a

3 Answers
3

C is indeed the correct answer. You need to understand the use of continue. continue statement is used inside loops. Whenever a continue statement is encountered inside a loop, control directly jumps to the beginning of the loop for next iteration, skipping the execution of statements inside loop's body for the current iteration.

continue

continue

So here in this case as the value of a reaches 4 after it gets printed the next statement if(a>3 && a<8) stands true hence continue, which skipped the rest code. In the next iteration again it does the same and will keep on doing it till infinite...

a

if(a>3 && a<8)

true

continue

Hope that explains you. Let me know if you want me to explain it further.

The answer is option C

When compiler encounters a continue statement then it ignore the whole code below it for that iteration.

In this case when a becomes 4 it was never incremented,so 4 gets printed infinite times after 3.

Once a is equal to 4, the continue statement is executed, which branches back to the beginning of the loop, bypassing the a++ statement. From this point on, a++ is never executed and the value of a never changes.

a

continue

a++

a++

a