On 6 July 2013 18:23, Namespace <rswhite4@googlemail.com> wrote:

That doesn't do what I want at all. The signature is still f(T) not
f(Unqual!T).

For non-const, const and immutable inout would do it.

void f(T)(inout T var)

Not if there's more than 1 argument, and 'inout T' is still not 'T'.