Why C# compiler is not throwing any error?

Question

Following is a piece of code:

    {
        int counter = 1;
        try
        {
            while (true) 
                counter*=2;
        }
        catch (Exception)
        {
            Console.WriteLine(counter);
            Console.ReadLine();
        }
    }

When I run this code, after a few iterations, value of 'counter' becomes 0. I don't understand why it is so?

How many is a "few iterations"? So it is not reaching the Exception part of your code when you are expecting it to? We need a bit of more an explanation in your question, but it does sound like an integer overflow — John Odom, May 20 '15 at 15:20
I'm not sure I understand the relation between the question's title and the body of the question. What is it you're asking about? Why it becomes 0? How does that relate to the C# compiler throwing an error? What would the error be about? — Borgleader, May 20 '15 at 15:21
`counter` overflows when the value assigned to it is larger than `int.MaxValue` and gets rolled back to `0` — Andrew Whitaker, May 20 '15 at 15:21
Why would the compiler throw an error? If the syntax is correct, the behaviour doesn't make any difference. — Joe, May 20 '15 at 15:21
very much unclear what are you asking. rethink on your question — HaveNoDisplayName, May 20 '15 at 15:22
@DStanley I figured :P. Just wanted OP to add more info to the question so we can be 100% sure with our answers/comments. — John Odom, May 20 '15 at 15:23

Dmitry Bychenko · Answer 1 · 2015-05-21T07:35:48.913

Use checked for the overflow exception to be thrown:

checked { 
  int counter = 1;

  try {
    while (true) 
      counter *= 2;
    }
  catch (Exception) { // Actually, on integer overflow
    Console.WriteLine(counter);
    Console.ReadLine();
  }
}

Edit: what's going on.

Fact: integer multiplication by 2 is equal to left shift by 1, i.e.

counter * 2 == counter << 1

In your case (let counter be represented as binary)

 00000000000000000000000000000001 // initial, just 1 
 00000000000000000000000000000010 // 1st itteration (shifted by 1)
 00000000000000000000000000000100 // 2nd itteration (shifted by 2)
 ...
 10000000000000000000000000000000 // 31st itteration (shifted by 31)

the next, 32nd itteration can cause either integer overflow or when unchecked just push the leftmost 1 out

 0000000000000000000000000000000 // 32nd itterartion, now we have 0

score 3 · Accepted Answer · answered May 20 '15 at 15:26

3

When counter reach the int.MaxValue then counter * 2 becomes a negative integer.

Then when counter reach int.MinValue then counter * 2 becomes 0.

Then on each iteration you have 0 * 2 = 0, no exception to throw.

answered May 20 '15 at 15:26

Actually it will go up to 1073741824, then `int.MinValue`, then 0. – juharr May 20 '15 at 15:34

Why C# compiler is not throwing any error?

2 Answers2