No implicit int -> short conversion in ternary statement

Question

short s;
s = (EitherTrueOrFalse()) ? 0 : 1;

This fails with:

error CS0266: Cannot implicitly convert type \'int\' to \'sho

Answer 1

I guess this has the same reason as this won't compile:

short s1 = GetShort1();
short s2 = GetShort2();
short s3 = s1 + s2;

I.e. that whenever short is used for something, it gets promoted to int.

Answer 2

The compiler has an implicit conversion from a constant expression to various primitive types (so long as the value is within the appropriate range), but here the expression isn't constant - it's just an int expression. It's pretty much the same as:

short s;
s = CallSomeMethodReturningInt32();

as far as the compiler is concerned.

There are two options - you could cast the whole expression, or cast each of the latter two operands:

short s = (EitherTrueOrFalse()) ? (short) 0 : (short) 1;

to make the overall expression type short. In this particular case, it's a pity that there isn't a numeric literal suffix to explicitly declare a short literal. Apparently the language designers did consider this, but felt it was a relatively rare situation. (I think I'd probably agree.)

The part about implicit constant conversions is from the C# 3.0 spec section 6.1.8:

6.1.8 Implicit constant expression conversions

An implicit constant expression conversion permits the following conversions:

• A constant-expression (§7.18) of type int can be converted to type sbyte, byte, short, ushort, uint, or ulong, provided the value of the constant-expression is within the range of the destination type.
• A constant-expression of type long can be converted to type ulong, provided the value of the constant-expression is not negative.

Answer 3

Because the cast is done by the compiler, not at runtime, I wouldn't call it an ugly cast, I would call it a complicated syntax:

s = (EitherTrueOrFalse()) ? (short)0 : (short)1;

I mean, this is the way it is written in C#, even if it looks ugly.

See this blog article. See Marc Gravell's answer on that question.