On Sunday, 17 May 2015 at 21:34:21 UTC, tcak wrote:
[code]
void test(D)( const D data ) if( is(D: shared(char[]) ) ) { }
void main() {
char[] text = new char[4];
text[0] = 'a'; text[1] = 'b'; text[2] = 'c'; text[3] = 'd';
auto t = cast( shared(const(char[])) )text[1..2];
test( t );
}
[/code]
Error Message:
template main.test cannot deduce function from argument types
!()(shared(const(char[])))
`t` is already shared(const(char[])), and `test` is expecting
const shared(char[]). Aren't they already same?
Per the template constraint, test needs a type D that implicitly
converts to shared(char[]). Even when the top const is removed
from typeof(t), there's still const in there:
shared(const(char)[]). That isn't implicitly convertible to
shared(char[]). Add const in the constraint and it works:
if( is(D: shared(const char[])) )
--
Together with this question, I want to ask whether there is a
way to check only being `shared`,
is(T == shared)
only being `const`,
is(T == const)
or only being `char[]`
I guess you mean to ignore any qualifiers of the array itself and
the element type.
We have std.traits.Unqual, but that only removes qualifiers from
the top level. As far as I know, we don't have anything like a
DeepUnqual which would allow you to write is(DeepUnqual!T ==
char[]). You could write one.
Or you can get into the more advanced versions of the
IsExpression:
is(T == E[], E) && is(Unqual!E == char)
of a template variable (D of `test` in this case)?
