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kdb

kdb: calling parse on a list of variables

I learned from here that

[...] in a parse tree a variable is represented by a symbol containing its name. Thus to distinguish a symbol or a list of symbols from a variable it is necessary to enlist that expression.

Given that, why does below expression evaluate to (enlist;`a;`b) instead of just `a`b?

Asking because it seems enlist[`a;`b]~`a`b is true.

q)parse"(a;b)"
enlist
`a
`b
Solution

  • I don't think you would ever want a parse tree to collapse two same-type values into a single uniform list, it would break the ability to eval the parse tree, e.g

    q)a:1
q)b:1
q)
q)eval parse"(a;b)"
1 1
q)eval `a`b
'type
  [0]  eval `a`b
       ^

    And secondly, (enlist;`a;`b) isn't the same as enlist[`a;`b] however the value of (enlist;`a;`b) is:

    q)value[(enlist;`a;`b)]~enlist[`a;`b]
1b

    So I guess it comes down to the nuanced differences between eval and value