CODEWITHSUNDEEP
c#
Brahim Boulkriat
Brahim Boulkriat

Reputation: 992

Overflow error when doing arithmetic operations with constants

I tried the following code :

int x, y;
x = y = int.MaxValue;

int result = x + y;

This code work fine and result will contain -2 (I know why).

But when doing this :

const int x = int.MaxValue;
const int y = int.MaxValue;

int result = x + y;

This will not compile because of overflow problem.Why ?

Upvotes: 12

Views: 265

Answers (1)

Jon Skeet
Jon Skeet

Reputation: 1503839

Because both x and y are compile-time constants, so is x + y. The compiler knows that the result will overflow, so it complains about it.

You can fix this by using an unchecked expression:

int result = unchecked(x + y);

From section 7.6.12 of the C# 5 specification - after listing +, -, / and *:

When one of the above operations produce a result that is too large to represent in the destination type, the context in which the operation is performed controls the resulting behavior:

  • In a checked context, if the operation is a constant expression (§7.19), a compile-time error occurs.
  • In an unchecked context, the result is truncated by discarding any high-order bits that do not fit in the destination type.

For non-constant expressions (expressions that are evaluated at run-time) that are not enclosed by any checked or unchecked operators or statements, the default overflow checking context is unchecked unless external factors (such as compiler switches and execution environment configuration) call for checked evaluation.

For constant expressions (expressions that can be fully evaluated at compile-time), the default overflow checking context is always checked. Unless a constant expression is explicitly placed in an unchecked context, overflows that occur during the compile-time evaluation of the expression always cause compile-time errors.

Upvotes: 13

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