Multiple Lexers from the same factory can't be interleaved

e.g.:

const factory = moo([['a', 'a'], ['b', 'b']])

const l1 = factory('ab')
const l2 = factory('ab')

l1.lex() // a
l2.lex() // b, should be a

[tjvr]

Perhaps add reset(input)

My thoughts exactly! 😃

I think I’m gonna rename it moo.compile, to make clear that building a lexer this way is expensive (I don’t want users to be calling moo() every time they want to parse something).

[nathan]

I think I measured it to have a slight cost?

I said “cheap”, not “free” 😃

Regardless, I’m still not sure what the motivation for factory() is.

Agreed. It seems like since the fastest thing to do is use a single Lexer and keep resetting it, the API should probably reflect that. Perhaps add reset(input) (or init, I’m not sure what to call it), which would do essentially the same thing as rewind(0); feed(input) but without the slicing/concatenation.

const lexer = moo({
  ws:      /[ \t]+/,
  rparen:  ')',
  keyword: ['while', 'if', 'else', 'moo', 'cows'],
  NL:      /\n/,
])

lexer.reset('while (10) cows\nmoo')
lexer.lex() // -> { name: 'keyword', value: 'while' }

lexer.reset('if (0) cats\ncattle')
lexer.lex() // -> { name: 'keyword', value: 'if'}