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.

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));

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?

On 09/26/2012 03:50 PM, monarch_dodra wrote:
> 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?

It is special syntax, not an == operator. is-expressions suppress errors
in the first argument only.

On 9/26/12 9:50 AM, monarch_dodra wrote:
> 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

Neither should work. The expression should be takeExactly(S.init, 1).

Andrei

On 09/26/2012 05:08 PM, Andrei Alexandrescu wrote:
> On 9/26/12 9:50 AM, monarch_dodra wrote:
>> 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
>
> Neither should work. The expression should be takeExactly(S.init, 1).
>
> Andrei

http://d.puremagic.com/issues/show_bug.cgi?id=8220
https://github.com/D-Programming-Language/dmd/pull/1007