On Tuesday, 19 July 2016 at 15:40:20 UTC, Lodovico Giaretta wrote:
You can find this out from the error, which says that you can't
expand an object of type `(Tuple!(int, int))`. Note the
surrounding parenthesis: they tell you that what you have is
not a Tuple, but an AliasSeq whose only membe
On Tuesday, 19 July 2016 at 15:36:42 UTC, Lodovico Giaretta wrote:
As you have to do `isTuple!(T[0])`, you also have to do
`x[0].expand`.
That's because T... works "as if" it was an array of types, and
x, being of type T, it works "as if" it was an array of values.
So you have to use an index i
On Tuesday, 19 July 2016 at 13:33:41 UTC, jmh530 wrote:
On Tuesday, 19 July 2016 at 07:23:52 UTC, John wrote:
auto bar(T...)(T x)
{
static if (T.length == 1 && isTuple!(T[0]))
return foo(x.expand);
else
return foo(x);
}
Hmm, this actually doesn't seem to be resolving my issue. I'
On Tuesday, 19 July 2016 at 07:23:52 UTC, John wrote:
auto bar(T...)(T x)
{
static if (T.length == 1 && isTuple!(T[0]))
return foo(x.expand);
else
return foo(x);
}
Hmm, this actually doesn't seem to be resolving my issue. I'm
still getting the error about not being able to expand
On Tuesday, 19 July 2016 at 07:23:52 UTC, John wrote:
auto bar(T...)(T x)
{
static if (T.length == 1 && isTuple!(T[0]))
return foo(x.expand);
else
return foo(x);
}
void main()
{
auto x = tuple(1, 2);
auto y = bar(x);
auto z = bar(x.expand);
writeln(
On Tuesday, 19 July 2016 at 01:22:01 UTC, jmh530 wrote:
import std.typecons : isTuple, tuple;
import std.stdio : writeln;
auto foo(T...)(T x)
{
T[0] y;
foreach (i, e; x)
{
y += e;
}
return y;
}
auto bar(T)(T x)
{
static if (isTuple
In the code below, is it possible to combine both bar functions
into one function (I just put some random foo function to get it
to work)?
When I try to do it, I get errors that
no property 'expand' for type '(Tuple!(int, int))'
I think it has something to do with T... being considered a
tuple
