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c#.netintegeroverflow
dreadwail
dreadwail

Reputation: 15410

Best way to handle Integer overflow in C#?

Handling integer overflow is a common task, but what's the best way to handle it in C#? Is there some syntactic sugar to make it simpler than with other languages? Or is this really the best way?

int x = foo();
int test = x * common;
if(test / common != x)
    Console.WriteLine("oh noes!");
else
    Console.WriteLine("safe!");

Upvotes: 54

Views: 52551

Answers (6)

Michael Petito
Michael Petito

Reputation: 13161

I haven't needed to use this often, but you can use the checked keyword:

int x = foo();
int test = checked(x * common);

Will result in a runtime exception if overflows. From MSDN:

In a checked context, if an expression produces a value that is outside the range of the destination type, the result depends on whether the expression is constant or non-constant. Constant expressions cause compile time errors, while non-constant expressions are evaluated at run time and raise exceptions.

I should also point out that there is another C# keyword, unchecked, which of course does the opposite of checked and ignores overflows. You might wonder when you'd ever use unchecked since it appears to be the default behavior. Well, there is a C# compiler option that defines how expressions outside of checked and unchecked are handled: /checked. You can set it under the advanced build settings of your project.

If you have a lot of expressions that need to be checked, the simplest thing to do would actually be to set the /checked build option. Then any expression that overflows, unless wrapped in unchecked, would result in a runtime exception.

Upvotes: 121

Jesse Williams
Jesse Williams

Reputation: 662

So, I ran into this far after the fact, and it mostly answered my question, but for my particular case (in the event anyone else has the same requirements), I wanted anything that would overflow the positive value of a signed int to just settle at int.MaxValue:

int x = int.MaxValue - 3;
int someval = foo();

try
{
   x += someval;
}

catch (OverflowException)
{
   x = int.MaxValue;
}

Upvotes: 0

Chinjoo
Chinjoo

Reputation: 2832

The best way is as Micheal Said - use Checked keyword. This can be done as :

int x = int.MaxValue;
try   
{
    checked
    {
        int test = x * 2;
        Console.WriteLine("No Overflow!");
    }
}
catch (OverflowException ex)
{
   Console.WriteLine("Overflow Exception caught as: " + ex.ToString());
}

Upvotes: 8

YourUncleBob
YourUncleBob

Reputation: 95

Old thread, but I just ran into this. I didn't want to use exceptions. What I ended up with was:

long a = (long)b * (long)c;
if(a>int.MaxValue || a<int.MinValue)
    do whatever you want with the overflow
return((int)a);

Upvotes: 5

JaredPar
JaredPar

Reputation: 755307

Try the following

int x = foo();
try {
  int test = checked (x * common);
  Console.WriteLine("safe!");
} catch (OverflowException) {
  Console.WriteLine("oh noes!");
}

Upvotes: 22

Alon Gubkin
Alon Gubkin

Reputation: 57139

Sometimes, the simplest way is the best way. I can't think a better way to write what you wrote, but you can short it to:

int x = foo();

if ((x * common) / common != x)
    Console.WriteLine("oh noes!");
else
    Console.WriteLine("safe!");

Note that I didn't remove the x variable because it'd be foolish to call the foo() three times.

Upvotes: 6

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