How can I write a LPeg pattern that never matches anything?

Question

Is there a way to create an LPeg pattern that always fails to match anything? I'm not talking about matching the empty string, I'm talking about a pattern that always fails so if you put it in an ordered choice it always will fall back to the second option.

The reason for this is that I am writing a small parser with LPEG and I wish I could write

 operators = empty_pattern + "==" + "~=" + "<=" + ">=" + "<" + ">"

instead of

 operators = lpeg.P("==") + "~=" + "<=" + ">=" + "<" + ">"

On lpeg.re you can also try `(&. !.)`, which means there is a next char and there is no next char, which is always false since these two scenarios are mutually exclusive. — greatwolf, Feb 14 '17 at 00:15

nobody · Accepted Answer · 2017-02-13T22:19:50.680

2

lpeg.P( false ) is the simplest way.

If the argument is a boolean, the result is a pattern that always succeeds or always fails (according to the boolean value), without consuming any input.

edited Feb 13 '17 at 22:19

answered Feb 13 '17 at 22:00

nobody

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Thanks! This is exactly what I needed. – hugomg Feb 14 '17 at 16:43

How can I write a LPeg pattern that never matches anything?

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