Please help me with colon : operator, I'm stuck on how it works. It works as an assignment, assignment through x+:1, global assignment/view ::, I/O 0:, 1:, to return value from the middle of the function :r, and to get an unary form of operator #:.
But what happend if one apply an adverb to it? I tried this way:
$ q
KDB+ 3.6 2019.04.02 Copyright (C) 1993-2019 Kx Systems
q)(+')[100;2 3 4]
102 103 104
q)(:')[x;2 3 4]
'x
[0] (:')[x;2 3 4]
^
q)(:')[100;2 3 4]
2 3 4
I expect evaluations in order: x:2, then x:3, then x:4. To get x:4 as a result. But I've got an error. And also :' works with a number 100 for some unknown reason.
What :' is actually doing?
q)parse "(:')[100;2 3 4]"
(';:)
100
2 3 4
Parsing didn't shed much light to me, so I'm asking for your help.
When modified by an iterator (also known as an adverb in q speak), : behaves just like any other binary operator. In your example
q)(:')[100;2 3 4]
2 3 4
an atom 100 is extended to a conformant list 100 100 100 and then : is applied to elements of the two lists pairwise. The final result is returned. It might look confusing (: tries to modify a constant value, really?) but if you compare this to any other binary operator and notice that they never modify their operands but return a result of expression everything should click into place.
For example, compare
q)+'[100; 2 3 4]
102 103 104
and
q)(:')[100;2 3 4]
2 3 4
In both cases an a temporary vector 100 100 100 is created implicitly and an operator is applied to it and 2 3 4. So the former is semantically equivalent to
(t[0]+2;t[1]+2;t[2]+4)
and the latter to
(t[0]:2;t[1]:2;t[2]:4)
where t is that temporary vector.
This explains why (:')[x;2 3 4] gives an error -- if x doesn't exist kdb can't extend it to a list.