Mixing signed and unsigned in C#

Question

In C and C++ mixing signed and unsigned types in expressions can be tricky since here

void foo(void)
{
 unsigned int a = 6;
 int b = -20;
 (a+b > 6) ? puts("> 6") : puts("<= 6");
}

output could (unexpectedly) be ">6" due to promotions. b will be promoted to unsigned. Do same things happen in C#? Is it also dangerous to mix signed and unsigned in C# expressions?

did you try it? `uint a = 6; int b = -20; Console.WriteLine(a + b > 6 ? ">6" : "<6");` — DLeh, Oct 26 '15 at 20:22
http://stackoverflow.com/questions/26368068/addition-of-int-and-uint — DLeh, Oct 26 '15 at 20:23
it's dangerous if you don't take possible issues into account. try `Console.WriteLine(typeof(a+b));` That will tell you the type of the result. — GreatAndPowerfulOz, Oct 26 '15 at 20:25
@Gread.And.Powerful.Oz that will rather get you a syntax error :) — Lucas Trzesniewski, Oct 26 '15 at 20:25
@DLeh: So in C# you don't get surprises because of such operations? that answer answers just one case — , Oct 26 '15 at 20:26
@LucasTrzesniewski:So in C# you don't get surprises because of such operations? that answer answers just one case — , Oct 26 '15 at 20:27
@LucasTrzesniewski and what syntax error would you expect to get? It's an object and WriteLine takes an object and will call ToString() on it. — GreatAndPowerfulOz, Oct 26 '15 at 20:27
@BenVoigt: Can I get unexpected results by mixing these types? — , Oct 26 '15 at 20:28
@DLeh: Actually that what you marked duplicate doesn't answer the question in the way I stated! — , Oct 26 '15 at 20:28
@Quser nothing is "unexpected" since every type promotion is well-defined in C#. When in doubt, simply use `var` (this is the equivalent of C++'s `auto`) — Lucas Trzesniewski, Oct 26 '15 at 20:28
@Quser: I don't recall the C# type-handling rule offhand. Type promotions are covered in the language spec. `((int)(a+b))` should be safe. — Ben Voigt, Oct 26 '15 at 20:29
@quser from the answer in the other thread: `The other implicit numeric conversions never lose any information. `, so you will be safe doing this. — DLeh, Oct 26 '15 at 20:29
@LucasTrzesniewski: In C++ you can see the result is clearly unexpected, so I was wondering how in C#? I have feeling this is not duplicate of the question linked — , Oct 26 '15 at 20:30
@Quser actually I'm pretty sure type promotion is well-defined in C++ too, but not necessarily to the same types as in C#. Most of C# code in the wild doesn't bother to use unsigned types, and just default to `int`, even when the value should intrinsically be unsigned. — Lucas Trzesniewski, Oct 26 '15 at 20:31
C++ can bring and make more surprises in such situations than C# — Schullz, Oct 26 '15 at 20:32
@LucasTrzesniewski: It is well defined but if you don't know promotion rules would you expect ">6" in my example? — , Oct 26 '15 at 20:32
@DLeh: I don't think this is exact dupe because this question could have a more general answer but you closed it — , Oct 26 '15 at 20:35
@Quser my stance on this is that you should know the language you're using, no excuses :-) In C++ I'd expect at least a compiler warning for this, and for me getting a warning means the code is broken. The duplicate question's answers contain relevant spec citations, and there's no more general answer than the detailed spec. — Lucas Trzesniewski, Oct 26 '15 at 20:36
@LucasTrzesniewski: "In C++ I'd expect at least a compiler warning for this, and for me getting a warning means the code is broken" - I don't think you will necessarily get warning, depends on compiler maybe — , Oct 26 '15 at 20:37
@DLeh: I'd advise to vote to reopen, maybe this one attracts more digestable answers — , Oct 26 '15 at 20:46
@Quser i will vote how i choose to vote, thank you very much. And I think that your bolded question "is it dangerous" is easily answered by the quote I mentioned before which specifically says that it is always safe. — DLeh, Oct 26 '15 at 20:47
@Quser that's not going to happen - your question is the textbook definition of a duplicate, and the answers in the other question are very good and complete. There's no need to duplicate that information here. — Lucas Trzesniewski, Oct 26 '15 at 20:50

Mixing signed and unsigned in C#

0 Answers0