C# Compiler Detecting Overflow?

Daniel Roberts

I am passing 0xffff0000 to a DirectX function that takes an int and interprets the int as a color (fully transparent red). The C# compiler says this is an error because it is auto-converting 0xffff0000 to a uint and saying it can't convert the value to an int (which is what the function takes). Is the C# compiler disallowing overflow? This wouldn't be a problem in C/C++ because the function looks at the byte content of the parameter anyway. Is there a solution to this?

By the way, is there a good resource (web or book) where I can find gory C# details like this? Something like Stroustrup's C++ book?

Thanks

3 Replies

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Alan18y ago
It seems to be the only way you can get it through the compiler. I wondered whether compiling with the /checked- switch (see link below) might enable you to avoid using unchecked but it only seems to affect runtime behaviour:

http://msdn.microsoft.com/en-us/library/h25wtyxf(VS.71).aspx
Daniel Roberts18y ago
Awesome. Thanks for the response. Excuse the stupid question but is using the unchecked function how you would recommend doing it, or is there another preferred way to deal with hexadecimal switches (like colors) in C#?
Thanks
Alan18y ago
The rule with integer literals in C# is that they're treated as type 'int' if they're within the int range, otherwise they're treated as type 'uint'.

However, you can force them to 'wrap' using the unchecked function:

int x = unchecked((int)0xffff0000);
Console.WriteLine(x); // -65536

The best place to look for these sort of details is the C# Specification :

http://msdn.microsoft.com/en-gb/vcsharp/aa336809.aspx