digitalmars.D.learn - Const is already there. It cannot deduce it
[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? -- Together with this question, I want to ask whether there is a way to check only being `shared`, only being `const`, or only being `char[]` of a template variable (D of `test` in this case)?
May 17 2015
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)?
May 17 2015