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docs: Add comment re expr_call duplication #4638

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docs: Add comment re expr_call duplication #4638

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65e8c44
Add comment re `expr_call` duplication
max-sixty Jun 20, 2024
fae0021
Merge branch 'main' into question-re-expr-call
max-sixty Jun 22, 2024
7acc477
Merge branch 'main' into question-re-expr-call
max-sixty Jun 22, 2024
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3 changes: 3 additions & 0 deletions prqlc/prqlc-parser/src/expr.rs
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Original file line number Diff line number Diff line change
Expand Up @@ -27,6 +27,9 @@ pub fn expr() -> impl Parser<TokenKind, Expr, Error = PError> + Clone {
.map(|x| x.to_string())
.map(ExprKind::Internal);

// This simply inlines `expr_call`. But if we call `expr_call` here, we get a
// stack overflow — possibly because this is a recursive parser?
// let nested_expr = pipeline(expr_call().boxed()).boxed();
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It's not exactly an inline:

  • the expr in this function is the reference to the "recursive" node,
  • the expr in expr_call() is the whole expr parser, reconstructed.

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To be fair the is the most confusing part of the parser (and prql cfg).

It stems from the fact that some locations is PRQL allow "expression that can have function calls at top-level" and other locations allow only "expression that don't have function calls at top-level".

For example:

  • let a = ____ is the first case,
  • join ____ (owner==id) is the second case. Here, the ___ cannot be expr with function calls at the top-level (for example from db.mytable) because that would also consume the last argument to join. It would parse as (join (from db.mytable (owner==id)) instead of (join (from db.mytable) (owner==id).

This stems from the fact that we don't require parenthesis for func calls or even commas.

let nested_expr = pipeline(lambda_func(expr.clone()).or(func_call(expr.clone()))).boxed();

let tuple = ident_part()
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