CODEWITHSUNDEEP
c#overflowexception
renil
renil

Reputation: 409

Why adding two large integers whose result greater than int.MaxValue doesn't throw overflow exception?

If we are using the following loop in a program, the loop never ends in C# 4.0

for (int i = 1; i <= int.MaxValue; i++)
{
}

This is because adding 1 to int.MaxValue (2147483647) will not result in an overflow exception, but results in -2147483648 (taking into consideration 32bit int and 2's compliment).

int i = int.MaxValue;
Console.WriteLine(i + 1);

It seems the behavior changed recently. See the question Arithmetic operation caused OverflowException .What could be the reason behind this change?

Upvotes: 5

Views: 3025

Answers (1)

Oded
Oded

Reputation: 498992

Overflow exceptions for integer (and other integral types) are only done in checked contexts.

So, this will cause an exception:

checked
{
 int i = int.MaxValue;
  Console.WriteLine(i + 1);
}

They are not set to do this by default as they are more expensive than simply overflowing.

From MSDN:

The checked keyword is used to explicitly enable overflow checking for integral-type arithmetic operations and conversions.

And:

Overflow checking can be enabled by compiler options, environment configuration, or use of the checked keyword.

This is not a recent change - C# has been like this from day one. What you see in the question is VB.NET code, which is by default in a checked context.

So, keeping to defaults, overflowing code in VB.NET will throw an exception, but identical code in C# will not.

Upvotes: 10

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