parcr (version 0.6.1)

zero_or_more: Repeated application of a parser

Description

Often, we want to assess whether a given structure can be successfully parsed through repetitive application of a parser p. This could involve testing the parser applied multiple times in succession or determining its capability to be applied zero or more times.

The subsequent functions are designed to address and evaluate these scenarios.

Usage

zero_or_more(p)
one_or_more(p)
exactly(n, p)
zero_or_one(p)
match_n(n, p)

Value

A parser.

Arguments

p

a parser.

n

a positive integer, including 0.

Pseudocode

zero_or_more(p):
  (p %then% zero_or_more(p)) %or% succeed(null)

one_or_more(p): p %then% zero_or_more(p)

exactly(n,p): count = 0 r = zero_or_more(p %using% F(x): count = count + 1; x)(x) if count == n then count = 0 r else fail()(x)

zero_or_one: exactly(1,p) %or% exactly(0,p)

match_n(n,p): if n==0 then F(x): succeed(list())(x) else if n==1 then p else (p %then% match_n(n-1, p))

where null is the empty vector.

Details

All these parsers with the exception of match_n exhibit greedy behavior striving to apply p as many times as possible. If the resulting count doesn't match the expected quantity, such as in the case of exactly(n,p) where p successfully parses more than n times, then the parser fails. In contrast, match_n(n,p) strictly applies p exactly n times, preventing any further application of p even if p could potentially be applied more often. Clearly, both functions will fail if p fails after less than n repetitions.

Examples

zero_or_more(literal("A"))(c("A", LETTERS[1:5]))
zero_or_more(literal("A"))(LETTERS[2:5])

one_or_more(literal("A"))(c("A", LETTERS[1:5])) # success
one_or_more(literal("A"))(LETTERS[2:5]) # failure

exactly(2, literal("A"))(c("A", LETTERS[1:5])) # success
exactly(2, literal("A"))(c(rep("A", 2), LETTERS[1:5])) # failure: too many "A"

zero_or_one(literal("A"))(LETTERS[2:5]) # success
zero_or_one(literal("A"))(LETTERS[1:5]) # success
zero_or_one(literal("A"))(c("A", LETTERS[1:5])) # failure

match_n(2, literal("A"))(c("A", LETTERS[1:5])) # success
match_n(2, literal("A"))(c(rep("A", 2), LETTERS[1:5])) # success