Lesson 24 · Evaluating expressions against a row

Evaluating expressions against a row

To run a query you must compute an expression's value for a given row. Today you build the evaluator for the two atoms - a literal yields itself, a column reference reads the row.

The goal

Evaluate a literal or column-reference expression against a row and its schema to produce a value.

Start here - the target

TO DO

Scenario: Evaluating atoms against a row

Giventhe schema (id INTEGER), (name TEXT) and the row [7, "alice"]

Whenthe column reference "name" is evaluated

Thenthe result is the text value "alice"

Andevaluating the literal 42 yields the integer value 42

Andevaluating an unknown column "zzz" reports an error

Background

Running a WHERE or a SELECT means asking, for each row, “what does this expression evaluate to here?” Evaluation answers that. It is a function over the expression tree, and the leaves are the base cases: a literal evaluates to itself (independent of the row), and a column reference evaluates by looking its name up in the schema to get an index, then reading that field from the row.

This is where the schema’s name-to-index lookup from lesson 3 finally pays off - it turns the name "name" into position 1 so the evaluator can pull the value out. A reference to a column that does not exist is an error caught right here. With the leaves evaluating, adding the interior nodes - comparisons and boolean operators - is a short step, and that is exactly the predicate a filter needs.

Make it work

func Eval(e Expr, row Row, s Schema) (Value, error) {
  switch e := e.(type) {
  case IntLit: return Value{Kind: KindInt, Int: e.Value}, nil
  case StrLit: return Value{Kind: KindText, Text: e.Value}, nil
  case ColRef: /* schema.IndexOf(e.Name) then row.Field(i) */
  }
}

CheckpointDONE

The evaluator computes literals and column references against a row. Commit and stop here.