views:

105

answers:

3

Hello all

I've a for loop which keeps incrementing an integer value till the loop completes. So if the limit n is a double variable and the incremented variable 'i' is an integer, i gets incremented beyond its limits.

 double total = 0;
 double number = hugetValue;
 for (int i = 1; i <= number; i++)
 {
    total = total + i;
 }
 return total;

What happens to 'i' if it exceeds its capacity? How the value of i changes? Will i get a runtime error?

Thanks

NLV

+3  A: 

Apologies if this seems rude, but you will learn far more by trying this yourself.

Edited: aha, so you did try it, and got unexpected results. As has been explained elsewhere C-like languages tend to quietly wrap integer arithmetic. That's actually quite a reasonable behaviour in general if the cost of checking for overflow is high. Once you know that this can happen one codes carefully, especially watching for the kind of construct in your example.

djna
Took it in the right sense. I actually tried it and was stuck in an infinite loop. When i debugged it that i saw the value of i to be negative. I was wondering why it dint throw an error and instead was continuing the loop. Any reasons for that?
NLV
There are two different issues here, why the conversion yields a negative number and why does the compiler not complain on the condition of the value being negative. The latest is the simpler: you could be modifying the loop variable inside the loop and making it negative. Detecting if this happens or not is too complex a problem for the compiler to try, so it just assumes that you know what you want (or you will find soon enough when you are stuck in your infinite loop)
David Rodríguez - dribeas
+1  A: 

Similar to the behaviour in C (where an int just wraps around from INT_MAX to INT_MIN), C# also wraps. Testing it in VS2008 with:

int x = 2147483647;
if (x+1 < x) {
    MessageBox.Show("It wrapped...");
}

will result in the message box appering.

If your hugetValue is greater than the maximum int value, then your loop will run forever because of this.

For example, if it's 2147483648, just as you think you're getting close to it, the int wraps around from 2147483647 back to -2147483648 and the loop just keeps on going.

paxdiablo
I wrote in C#. Sorry for not mentioning that. I dint understand your answer.
NLV
@NLV: Some goes for C#.
leppie
@NLV: I have edited the question to include the C# tag. Note that in many questions the language is important (some allow the conversion implicly, some will not allow it at compile time). When making a question the more information you provide --within reasonable terms-- the better.
David Rodríguez - dribeas
Thank you for the tip David. My bad.
NLV
@paxdiablo @NLV Also you can use `Int32.MaxValue` http://msdn.microsoft.com/en-us/library/system.int32.maxvalue.aspx
abatishchev
+1  A: 

If you want an exception, either supply the checked compiler option, or use the checked construct provided in C#.

leppie