Best way to handle Integer overflow in C#?

Question

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!");

• Mitch Wheat: best way is to prevent in first place
• dreadwail: Sure, but that's a different question from the one presented here. Handling it and preventing it are seperate (related of course) discussions.

Answer 1

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.

• KFL: Why is the default behavior is unchecked? Is there any performance concerns on using checked?
• Michael Petito: @KFL "Because checking for overflow takes time, the use of unchecked code in situations where there is no danger of overflow might improve performance. However, if overflow is a possibility, a checked environment should be used." See msdn.microsoft.com/en-us/library/a569z7k8.aspx. I also believe that the default is unchecked because it parallels the C / C++ behavior.
• user472308: There are cases where overflow is useful, for instance the TCP sequence number. An unsigned int just wraps around to zero, so all that's needed is the increment operator.
• user5528169: Does unchecked produce same results in case of overflow on all machines??
• Saurabh Kumar: Updated url reference for 'checked' - msdn.microsoft.com/en-us/library/74b4xzyw(v=vs.140).aspx

Answer 2

Try the following

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

Answer 3

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());
}

• dreadwail: You need to format your code (insert 4 spaces prior to it and the site will automatically pick up the formatting and syntax highlighting.

Answer 4

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.

• Michael Petito: Now I wonder if the compiler would ever optimize that expression away entirely?
• Cobusve: USefull, unless of course you are planning on using the result (x*common), in that case your shortening would require the calculation to be done twice ...
• Neil: What if common == 0?

Answer 5

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);

• Ray Cheng: This won't work. int.MaxValue + 1 gives you -2147483648 or int.MinValue. so a can never be larger than MaxValue or smaller than MinValue. Think of numbers as a line like this: -2147483648, -2147483647, ... 0 ... 2147483646, 2147483647 when you go over the max value, it flows back to the smallest.
• hammett: you missed the fact that he/she converted to long before checking against the bounds of the int. so it would work as long it doesnt overflow the long.

Answer 6

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;
}

• Paul Knopf: The exception wouldn't get thrown unless you compile with /checked, or use the checked keyword, correct?