A library for parsing Backus–Naur form context-free grammars.

Add to your Cargo.toml:

[dependencies] bnf = "0.6"

The following grammar from the Wikipedia page on Backus-Naur form exemplifies a compatible grammar. (*Note: parser allows for an optional ';' to indicate the end of a production)

 <postal-address> ::= <name-part> <street-address> <zip-part> <name-part> ::= <personal-part> <last-name> <opt-suffix-part> <EOL> | <personal-part> <name-part> <personal-part> ::= <initial> "." | <first-name> <street-address> ::= <house-num> <street-name> <opt-apt-num> <EOL> <zip-part> ::= <town-name> "," <state-code> <ZIP-code> <EOL> <opt-suffix-part> ::= "Sr." | "Jr." | <roman-numeral> | "" <opt-apt-num> ::= <apt-num> | "" 

When parsing grammar text (e.g. [str::parse] or [Grammar::parse_from]), the parser accepts two shortcuts:

Group alternatives so they act as one unit in a sequence.

Without parentheses, | binds loosely. This rule:

<s> ::= <a> | <b> <c> 

means "<a> or <b> <c>". So "a" matches, and "b c" matches, but "a c" does not.

With parentheses, you get "(a or b) then c":

<s> ::= (<a> | <b>) <c> 

So only "a c" and "b c" match.

Zero or one of the grouped alternatives (like ? in regex).

<word> ::= <letter> [<digit>] 

means: a letter, optionally followed by a digit. Both "x" and "x1" match.

Equivalent long form without extended syntax:

<word> ::= <letter> <opt-digit> <opt-digit> ::= <digit> | "" 

Groups and optionals are normalized into a grammar that uses only plain nonterminals and terminals: each group or optional is turned into a fresh internal nonterminal (e.g. __anon0, __anon1). The public [Term] type has only [Term::Terminal] and [Term::Nonterminal]; parsing and generation use this normalized form.

Round-trip: Formatting a grammar (e.g. format!("{}", grammar)) does not preserve ( ) or [ ]; the result uses __anon* names. Re-parsing yields an equivalent grammar.

Empty groups or optionals — () or [] with nothing inside — are invalid; at least one alternative is required.

Take the following grammar for DNA sequences to be input to this library's parse function.

<dna> ::= <base> | <base> <dna> <base> ::= "A" | "C" | "G" | "T" 

The output is a Grammar object representing a tree that looks like this:

Grammar ├── <dna> ::= │ ├── <base> │ └── <base> <dna> └── <base> ::= ├── "A" ├── "C" ├── "G" └── "T" 

Once the Grammar object is populated, to generate a random sentence from it call the object's generate function. grammar.generate(). For the above grammar you could expect something like TGGC or AG.

If the generate function can't find a production for a nonterminal it tries to evaluate it will print the identifer as a nonterminal, i.e. <identifier>.

The generate function will return an error if it detects an infinite loop caused by a production such as <PATTERN> ::= <PATTERN>.

use bnf::Grammar; let input = "<postal-address> ::= <name-part> <street-address> <zip-part>   <name-part> ::= <personal-part> <last-name> <opt-suffix-part> <EOL>  | <personal-part> <name-part>   <personal-part> ::= <initial> '.' | <first-name>   <street-address> ::= <house-num> <street-name> <opt-apt-num> <EOL>   <zip-part> ::= <town-name> ',' <state-code> <ZIP-code> <EOL>   <opt-suffix-part> ::= 'Sr.' | 'Jr.' | <roman-numeral> | ''  <opt-apt-num> ::= <apt-num> | ''"; let grammar: Result<Grammar, _> = input.parse(); match grammar { Ok(g) => println!("{:#?}", g), Err(e) => println!("Failed to make grammar from String: {}", e), }

use bnf::Grammar; let input = "<dna> ::= <base> | <base> <dna>  <base> ::= 'A' | 'C' | 'G' | 'T'"; let grammar: Grammar = input.parse().unwrap(); let sentence = grammar.generate(); match sentence { Ok(s) => println!("random sentence: {}", s), Err(e) => println!("something went wrong: {}!", e) }

use bnf::Grammar; let input = "<dna> ::= <base> | <base> <dna>  <base> ::= 'A' | 'C' | 'G' | 'T'"; let grammar: Grammar = input.parse().unwrap(); // Create a parser from the grammar (validates all nonterminals are defined) let parser = grammar.build_parser().unwrap(); let sentence = "GATTACA"; let mut parse_trees = parser.parse_input(sentence); match parse_trees.next() { Some(parse_tree) => println!("{}", parse_tree), _ => println!("Grammar could not parse sentence"), }

By default, parse_input implicitly starts from the first rule. To match another rule, parse_input_starting_with can be used:

use bnf::{Grammar, Term}; let input = "<dna> ::= <base> | <base> <dna>  <base> ::= 'A' | 'C' | 'G' | 'T'"; let grammar: Grammar = input.parse().unwrap(); // Create a parser from the grammar (validates all nonterminals are defined) let parser = grammar.build_parser().unwrap(); let sentence = "G"; let target_production = Term::Nonterminal("base".to_string()); let mut parse_trees = parser.parse_input_starting_with(sentence, &target_production); match parse_trees.next() { Some(parse_tree) => println!("{}", parse_tree), _ => println!("Grammar could not parse sentence"), }