Re: is expression on templated types

[simendsjo]

On 25.06.2011 16:32, Simen Kjaeraas wrote:
On Sat, 25 Jun 2011 15:57:03 +0200, simendsjo
<simen.end...@pandavre.com> wrote:
I have a templated struct, and I'd like to check if a given type is
this struct. I have no idea how I should write the is expression..

struct S(int C, int R, T) {
T[C][R] data;
}
template isS(T) {
enum isS = is(T : S); // How can I see if T is a kind of S no matter
what values S is parameterized on?
}
unittest {
static assert(isS!(S!(1,1,float)));
static assert(!isS!(float[1][1]));
}

template isS(T) {
static if ( is( T t == S!(C,R,U), int C, int R, U ) ) {
enum isS = true;
} else {
enum isS = false;
}
}

This works, but I thought the following had been added. Apparently not:

template isS(T) {
static if ( is( T t == S!U, U... ) ) {
enum isS = true;
} else {
enum isS = false;
}
}

As for the weird way to write this, it is caused by some variants of the
isExpression not being available outside static if.


Thanks - works like a charm.
The is expression is quite a complex beast though.
Has anyone written any articles on it, or some more examples than the 
documentation?
Think I need many examples before I grok it.