Hello. I was trying to do something like this:

ubyte code = to!ubyte(spec, 6) + 16;

and got an error saying:

cannot implicitly convert expression (cast(int)to(spec, 6) + 16) of type int to ubyte

Looking at http://dlang.org/spec/lex.html#IntegerLiteral, sure enough 16 is specified to be inferred as an `int`.

I thought that integer literals used to be inferred as the smallest integral type that can fit them – am I mistaken?

-- 
Shriramana Sharma, Penguin #395953

They are promoted to int in arithmetic operations unless compiler can prove the value doesn't exceed its range.

On Monday, 14 December 2015 at 13:33:41 UTC, Shriramana Sharma wrote:
> ubyte code = to!ubyte(spec, 6) + 16;

That's not an integer literal... that's a runtime value of ubyte plus an integer literal.

Since the ubyte is the result of a runtime function, the compiler doesn't know what it will be and thinks it could be anything from 0 to 255, inclusive.

240 + 16 = 256 = too big to fit in a ubyte, to it requires a cast.

ubyte code = 16; // this would work

ubyte code = 239 + 16; // this too

but yours won't because to!ubyte(spec, 6) might just be > 240.

Adam D.  Ruppe wrote:
> but yours won't because to!ubyte(spec, 6) might just be > 240.

Thanks for that explanation. That's clear now.

-- 
Shriramana Sharma, Penguin #395953