Can someone please explain this code!?

Question

"What is the output of the following code?

int f = 1, i = 2;
while (++ i < 5) {
f* = i
System.out.println(f); "

I have copied this code from one of the quiz questions, and guessed the answer. The correct answer is 12.

(I understand I have left out the closing curly brackets/braces ( } ), as i dont remember where they were in the question. 😜

So if anyone can help i would really appreciate it.

Thanks in advance!! 👍🏻

Answer 1

int f = 1, i = 2;
while (++i < 5) { //increments i, then checks if it it lower than 5
f *= i; //multiplies f by i
}
System.out.println(f); //prints f

The output of this code is 3*4 = 12.

Answer 2

@Ahmad: ++i increments i by 1, so when we do the first loop, i is already equal to 3. So f is multiplied by i and becomes 1*3=3.

f *= i is a shorter synthax for f = f * i by the way.

Answer 3

int f = 1, i = 2;
while (++i < 5) { //increments i, then checks if it it lower than 5
, so i=3
f *= i; //multiplies f by i
(f=f*i) now f =1*3, therefore f=3
//then loop again executes as i=4,therefore f=3*4
}
System.out.println(f); //prints f

Answer 4

3*4 =12
the last value for i is 4 as 5 is not less thann 5

Answer 5

12

Answer 6

Thanks so much!! That totally makes sense now!! 😀👍🏻

Answer 7

thanks all you are heros in C++

Answer 8

Wait. How the f became three?

Answer 9

thanks!

Answer 10

int f = 1, i = 2;
while (++ i < 5) {
f = f*i;
System.out.println(f); "

Answer 11

12

Answer 12

output is 12

Answer 13

Why i execute to 4

Answer 14

Why we do 3*4 again

Answer 15

youre welcome

Answer 16

What is the output of the following code?
int f=1, i=2;
while(++i<5) {
f*=i;
}
cout<<f;

answer :
12

Answer 17

3 I--> 4 => 12 !!!

Answer 18

12

Answer 19

12

Answer 20

12