Testing for template argument being result of takeExactly

[monarch_dodra]

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?