int a=2; int b=3; while(b--)++a; cout<<a; Why does this code print 5?

Answer 1

for any integral values not equal to zero inside the while condition, the conditions holds good and it prints out,but when b-- becomes zero the condition becomes false and the final value of a that is 5 is printed due to
the following iterations
1st iteration b remains 3(post increments) and a becomes 3(pre increment)
2nd iteration b becomes 2 and a becomes 4
3rd iteration b becomes 1 and a becomes 5
------- ------- -------- -------- --------
AND NOW FINALLY B BECOMES ZERO CONDITION BECOMES FALSE AND THE FINAL VALUE OF A =5 IS PRINTED OUT :)

Answer 2

In short, any int that is greater than one is considered true. If it is 0 or less, it is considered false. As we know, while(true) is an infinite loop, while(false) will not run.

Inside of the parameters we have --b, it is important to know that this is done BEFORE the while loop checks if the condition is true or not. So as soon as b reaches 0, a will be the value in which b started with.

On the next iliteration, b-- is done again (hence why b = -1 if you print it out), and sees that it evaluates to false, thus the loop ends.

Bit of a silly loop to be honest and got my head in a muddle, but I learnt something new today, thanks :)

Answer 3

Cohen: number 0 is false, every others true.

Answer 4

So why when b = -1, the while loop stops? -1 would be true? This question confused me but I'm learning at the same time so it's great.