Output

Question

Int I=3;
If(!I)
I++;
I++;
If(I==3)
I+=2;
I+=2;
Printf("%d", I);

The output is 6 but how?

Answer 1

Lets format your code to understand more easily:

int l = 3;

if (!l) { // False. Doesn't execute.
l++;
}

l++; // l = 4

if (l==3) { //Doesnt execute either cause l=4 now.
l += 2;
}

l += 2; // 4 + 2
printf( %d", l);

See? When you doesn't use brackets for an if, else, for, while etc..., Only the [ONE] instruction immediately after it gets executed.
First you assign 3 to l. Then you check if l doesn't has a Truthy value, which is true because e is a true value, and the ! operator inverts the result which returns false to the if test. Then you increment by one and checke if l == 3, which is false cause at that point l is equal to 4, so it skips the if block and executes the last statement: l += 2.
4 + 2 = 6.

Answer 2

Achintya Raj

Here i is declared so !i will be false and if condition will not work only one i++ will work.

So here i will be 4

Now again i == 3 will be false so Only one i += 2 will work and i will be 6

Answer 3

Achintya Raj First i is 3 then it is increment & you add 2 then it becomes 6

Answer 4

But what is the reason

Answer 5

Thanks

Answer 6

Such an easy question I didn't recognise

Answer 7

Thanks for the ans