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"),
}