Editing Parsing expression grammar (section)

=== More examples ===
The following recursive rule matches standard C-style if/then/else statements in such a way that the optional "else" clause always binds to the innermost "if", because of the implicit prioritization of the '/' operator. (In a [[context-free grammar]], this construct yields the classic [[dangling else|dangling else ambiguity]].)
<syntaxhighlight lang="peg">
S ← 'if' C 'then' S 'else' S / 'if' C 'then' S
</syntaxhighlight>

The following recursive rule matches Pascal-style nested comment syntax, {{code|(* which can (* nest *) like this *)}}. Recall that {{code|.|peg}} matches any single character.
<syntaxhighlight lang="peg">
C     ← Begin N* End
Begin ← '(*'
End   ← '*)'
N     ← C / (!Begin !End .)
</syntaxhighlight>

The parsing expression {{code|2=peg|foo &(bar)}} matches and consumes the text "foo" but only if it is followed by the text "bar". The parsing expression {{code|2=peg|foo !(bar)}} matches the text "foo" but only if it is ''not'' followed by the text "bar".  The expression {{code|2=peg|!(a+ b) a}} matches a single "a" but only if it is not part of an arbitrarily long sequence of a's followed by a b.

The parsing expression {{code|2=peg|('a'/'b')*}} matches and consumes an arbitrary-length sequence of a's and b's. The [[Formal grammar#The syntax of grammars|production rule]] {{code|2=peg|S ← 'a' ''S''? 'b'}} describes the simple [[context-free language|context-free]] "matching language" <math> \{ a^n b^n : n \ge 1 \} </math>.
The following parsing expression grammar describes the classic non-context-free language <math> \{ a^n b^n c^n : n \ge 1 \} </math>:
<syntaxhighlight lang="peg">
S ← &(A 'c') 'a'+ B !.
A ← 'a' A? 'b'
B ← 'b' B? 'c'
</syntaxhighlight>