(Testing own parser to infer grammar) What would . tag | = "foo" mean?

[amn]

Mind you, the result seems different than if the white-space between the | and the = were removed. Without the white-space, I understand there's the |= operator, which has the effect of assigning "foo" as the value of the tag (using context of either the tag, if that even makes sense, or ., at the right hand side of the assignment).

With the white-space there, however, I don't understand what kind of expression I wrote, and the -v switch to yq doesn't seem to help me much, except some indication that it is no longer the equivalent of the |= operator being used, but rather the pipe (|) operator and the plain assignment (=) operator.

Can someone please shed some light on the difference between the two expression variants (with and without white-space), and what the one with the white-space between the | and the =, means? I understand also that some explicit parentheses might make it clear, but I am also not sure about the arity of the different operators in use here. E.g. if I were to rewrite the expression as (. tag |) = "foo" I believe it's equivalent of the variant where the parentheses were removed (without any other change, in particular to the white-space), but that doesn't make it much more clear what kind of parse tree or expression I am looking at.

[mikefarah]

So with no space |= is a single operator, and the expression tree looks like:

- |=
   - .tag (lhs argument)
   - "foo"(rhs argument)

With an (invalid) expression like ".tag | = "foo"' - yq will try to build an expression tree. Now each operator has an implied precedence (kind of like a maths expression, you evaluate * and / before + and -. You can see the precedence values in lib.go for what it's worth). In this case, = has a higher order of precedence than | and so it will build an invalid expression tree like this:

- =
   - | (LHS argument of "=")
      -  .tag (LHS argument of "|")
   - "foo" RHS argument of "="

And then it discovers the "|" operator only has one argument, whereas it needs two.

[amn]

Thanks. I think I was asking specifically about the variant . tag | = "foo" (space between . and tag)? It isn't an invalid expression here (yq v4.33.3). Anyway, I suppose the syntactical correctness here generally hinges on the simple fact that both | and = are binary operators -- requiring two operands, one on each side of the operator with an infix notation (that yq uses)?

[mikefarah]

Yep 👍

…

________________________________ From: Armen Michaeli ***@***.***> Sent: Friday, May 5, 2023 3:15:20 AM To: mikefarah/yq ***@***.***> Cc: Mike Farah ***@***.***>; Comment ***@***.***> Subject: Re: [mikefarah/yq] (Testing own parser to infer grammar) What would `. tag | = "foo"` mean? (Discussion #1657) Thanks. I think I was asking specifically about the variant . tag | = "foo" (space between . and tag)? It isn't an invalid expression here (yq v4.33.3). Anyway, I suppose the syntactical correctness here generally hinges on the simple fact that both | and = are binary operations -- requiring two operands, one on each side of the operator with an infix notation (that yq uses)? — Reply to this email directly, view it on GitHub<#1657 (reply in thread)>, or unsubscribe<https://github.com/notifications/unsubscribe-auth/AAIZHNI3QLCYWVRJIVM3PV3XEPP2RANCNFSM6AAAAAAXU3WHYE>. You are receiving this because you commented.Message ID: ***@***.***>