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.
"is(R == typeof(takeExactly(R, 1)))"
Makes sure they are the exact same type, and
"is(typeof(takeExactly(R, 1)))"
is a short-circuit, to prevent the second test from error'ing if
takeExactly!R is not legal.
