On Friday, 15 July 2016 at 15:04:22 UTC, Devin Hill wrote:
to the condition. It works pretty well! Granted, it doesn't
allow for calling it in two ways like a variadic version would
have:
foo(1, 2, 3) // works with this setup
foo([1, 2, 3]) // doesn't, but would only be occasionally
On Friday, 15 July 2016 at 05:23:15 UTC, Basile B. wrote:
even better:
template sameType(T...)
{
import std.meta;
static if (!T.length)
enum sameType = false;
else
enum sameType = NoDuplicates!T.length == 1;
}
Yeah, that's basically what I ended up doing, but
On Friday, 15 July 2016 at 04:38:03 UTC, Basile B. wrote:
two obvious:
- recursive template.
- staticIota in a foreach. (aliasSeqOf!(iota(1, T.length))
even better:
template sameType(T...)
{
import std.meta;
static if (!T.length)
enum sameType = false;
else
enum
On Friday, 15 July 2016 at 04:31:08 UTC, Devin Hill wrote:
Thanks, that way of doing it does work. I guess that means
there's no easy way to make sure all T are the same type
without a template constraint?
Yes, immediatly, now, I think that a contraint has to be used.
But you have several
On Friday, 15 July 2016 at 04:08:19 UTC, Basile B. wrote:
With D style variadics it works, you can build the array from
the list and have a static array:
=
void foo(T...)(T t)
{
T[0][T.length] tt = [t]; // T[0] is the type
writeln(tt); // [1,2,3]
static
On Friday, 15 July 2016 at 03:43:49 UTC, Devin Hill wrote:
Hi everyone,
I have a struct template which takes an integer n, and then has
a constructor taking that many arguments of type long, which
looks like:
struct Struct(int n) {
this(long[n] nums...) { /* stuff */ }
}
This works and
Hi everyone,
I have a struct template which takes an integer n, and then has a
constructor taking that many arguments of type long, which looks
like:
struct Struct(int n) {
this(long[n] nums...) { /* stuff */ }
}
This works and lets me change n for each instantiation, but I
wanted to
