On Sunday, 7 July 2013 at 11:07:44 UTC, Marco Leise wrote:
Am Sat, 6 Jul 2013 20:24:41 +1000
schrieb Manu <turkey...@gmail.com>:
On 6 July 2013 18:23, Namespace <rswhi...@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'.
Hey, inout looks like a neat solution! And it only creates one
template instance no matter how many arguments of type
inout(T) you use!
const would have the same effect:
void f(T)(const T var) { ... }
...but then you can't mutate var.
