All right. I think I'm starting to understand the reasoning that went into
the spec as written. Thank you for giving me the necessary pointers.
Now, that leads me inevitably to the obvious followup question: Is there a
way to run "my" version of this operator?
Regards,
Elias
On 21 March 2014 21
On 21 March 2014 13:39, Elias Mårtenson wrote:
> On 21 March 2014 19:55, Jay Foad wrote:
>> Associativity matters because the last item of the result is 3>1>2,
>> which means 3>(1>2). But your explanation would lead to (3>1)>2 as the
>> last item.
>
>
> You are right, but that can be addressed by
On 21 March 2014 19:55, Jay Foad wrote:
> >\3 1 2
> 3 1 1
>
> I think your explanation of scan would lead to the result 3 1 0.
>
> Associativity matters because the last item of the result is 3>1>2,
> which means 3>(1>2). But your explanation would lead to (3>1)>2 as the
> last item.
>
Yo
On 21 March 2014 11:48, Elias Mårtenson wrote:
> OK, I understand your explanation, and I have to say that the spec seems to
> be a bit over-specified here. My (simplistic?) understanding of the
> backslash operator was that it simply applied the given function on the
> first two elements, add the
On 19 March 2014 22:25, Kacper Gutowski wrote:
> On 2014-03-19 21:22:04, Elias Mårtenson wrote:
> > This can't possibly be correct?
>
> But it is!
> Let me just quote relevant excerpt of ISO 13751:
>
OK, I understand your explanation, and I have to say that the spec seems to
be a bit over-specif
On 20 March 2014 11:31, Kacper Gutowski wrote:
> On 2014-03-19 23:16:32, Elias Mårtenson wrote:
> > I see. So the way I see it, I should actually use 2 OP/X instead? I
> seems to
> > actually do what I expected.
>
> It if does what you need then sure, but it's a very different thing
> than scan:
On 2014-03-19 23:16:32, Elias Mårtenson wrote:
> I see. So the way I see it, I should actually use 2 OP/X instead? I seems to
> actually do what I expected.
It if does what you need then sure, but it's a very different thing
than scan:
,\'abcd'
a ab abc abcd
2,/'abcd'
ab bc cd
-
I see. So the way I see it, I should actually use 2 *OP*/X instead? I seems
to actually do what I expected.
Regards,
Elias
On 19 March 2014 22:25, Kacper Gutowski wrote:
> On 2014-03-19 21:22:04, Elias Mårtenson wrote:
> > This can't possibly be correct?
>
> But it is!
> Let me just quote rele
The following code returns the expected results:
* {⍺+⍵}\⍳10*
1 3 6 10 15 21 28 36 45 55
However, if I also print all the calls, I get a lot more evaluations of the
function than expected:
*{(⍺+⍵)⊣⎕←('[',(⍕⍺),',',(⍕⍵),']')}\⍳10*
[1,2]
[2,3]
[1,5]
[3,4]
[2,7]
[1,9]
[4,5]
[3,9]
[2,12]
