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@jemoka / Jemoka Knowledge Base / wiki/concepts/building_a_parser.md
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--- title: "top-down parsing" type: concept related: [Building A Parser] source: https://www.jemoka.com/posts/kbhbuilding_a_parser/ confidence: high status: active --- recursive descent parsing Given the grammar: E -> T | T + E T -> int | int * T | (E) And suppose our token stream is: (int5) We will start top-level non-terminal character E, and will try to apply the rules for E in order… (this is exactly how you imagine one builds a parser via recursive backtracking) when does it break? Consider S -> S a, this becomes S -> S1 a. Notice it will break into: bool S1() { return S() && term(a); } bool S() { return S1(); } This is what we call a left recursive grammar left recursive grammar That is, the grammar keep recurses to the left without termination. fixing left recursive grammars for top-down parsing For: S -> S a1 | ... | S an | b1 | ... | bm We write: S -> b1 S' | ... | bm S' S' -> a1 S' | ... | an S' | nothing where nothing is the empty string. example For the left recursive grammar S -> S a | b We can refactor as S -> B S' S' -> a S' | nothing where nothing is \(\epsilon\) example! We will define boolean functions to check the string for matches. // gobble up the token string pointer bool term(TOKEN tok) { return *next++ == tok; } Let’s now match each of our productions. bool E1() { return T(); } bool E2() { return T() && term(PLUS) && E(); } bool E() { TOKEN *save = next; // that is, if E1 matches, return immediately // otherwise, we return our saved pointer in text (i.e. roll what what // checking E1 could goble), and roll back return ( E1() || (next=save, E2()); ) } bool T1() { return term(INT) }; // ... you get the point