On Wednesday, 26 September 2012 at 13:19:13 UTC, Timon Gehr wrote:
On 09/25/2012 08:41 AM, monarch_dodra wrote:
On Monday, 24 September 2012 at 22:13:51 UTC, Timon Gehr wrote:
On 09/24/2012 09:41 AM, monarch_dodra wrote:
> [SNIP]
I don't think this does what you think it does. The 'is(R r)'
declares
r to be an alias for R. So 'r' is a type in that code snippet.
Darn :(
Also, is(typeof(takeExactly(R, 1))) && is(R ==
typeof(takeExactly(R, 1)))
can be written in a more compact way as
is(typeof(takeExactly(R, 1)) == R)
Technically, no: That was my first try, and as mentioned in
the first
reply, this returns true when the types of takeExactly and R
are equal
comparable, but does not make sure they are the actually the
same types.
...
I assume you messed up the parentheses.
is(typeof(takeExactly(R, 1) == R)) // test for equal-comparable
is(typeof(takeExactly(R, 1)) == R) // test for type identity
alias float flt;
static assert(is(typeof(1==flt)));
static assert(!is(typeof(1)==flt));
Yes sorry. I miss read that.
Kind of weird though, could you explain why:
struct S{};
is(typeof(takeExactly(S, 1)) == S) //false
is(S == typeof(takeExactly(S, 1))) //Error: template
std.range.takeExactly does not match any function template
declaration
I was under the understanding that == was a commutative
operation. Not the case in an is block?
