I was reading the spec on outer join to see if I could
finally fully understand how it works.
For reference, the spec refers to the syntax as follows:
X a.b Y
The function a is referenced twice in the description for
this function, both times in the form a/ (i.e. with the reduce
operator applied).
I then noted that the J version of this function does not
apply the reduce operator. I.e. in J, you have to do:
X a/.b Y
Now, am I correct in my assumption that with this change,
there would be no need for the ∘ (null function)? I.e. if APL
behaved like J, would the following achieve an outer join?
X ⊣.× Y
I'm still not sure I fully understand the join operator to
the point where I can answer this question.
Regards,
Elias