Daniel Keep wrote:
Michel Fortin wrote:
On 2008-12-31 21:50:53 -0500, Daniel Keep
<daniel.keep.li...@gmail.com> said:
Disclaimer: I'm not an expert on compilers. Plus, I just got up. :P
The key is that the parser has to know what "MyClass" means before it
can figure out what the "?" is for; that's why it's
context-dependant. D avoids this dependency between compilation
stages, because it complicates the compiler. When the parser sees
"MyClass", it *doesn't know* that it's a type, so it can't
distinguish between a nullable type and an invalid ?: expression.
As far as I know, this can't be done with pointer declarations. Read
this:
a * b;
Is it a multiplication or a declaration of a pointer to type "a"? You
don't know until you resolve the identifiers. It's the exact same
situation for using "?" to denote nullable.
At least, I think that's how it works; someone feel free to correct
me if it's not. :P
I belived the same for some time too, then found the above rebutal.
I don't think so; I believe that unary vs. binary operators are
unambiguous. "a * b" is "a multiply b" because it's "(expr) * (expr)".
It's pointer dereference when it's "* (expr)".
I keep in mind is that "(expr) < (expr) >" is ambiguous, but "* (expr)"
isn't.
Like I said, I'm not an expert, so take what I say with a grain of salt.
-- Daniel
It's about a pointer declaration, not a pointer dereference, a la "int * p;"
Happy New Year everyone!
Andrei
