Parsing a phrase with Sprache(Words seperated by spaces)

Question

I'm attempting to write a parser in Sprache that will parse a phrase

The basic rule is that it should include words seperated by a single space, with both the first and last character of the string being a space.

I would expect to call something like the following:

string phrase = PhraseParser.Parse("         I want to return up to this point        ");

And have the resulting string be "I want to return up to this point".

I have tried numerous implementations with none quite doing it for me.

Update Thanks to @PanagiotisKanavos, the trick would be to use the .Then() operator. The following words:

public static Parser<string> WordParser =
        Parse.Letter.Many().Text().Token();

public static Parser<string> PhraseParser =
        from leading in Parse.LetterOrDigit.Many().Text()
        from rest in Parse.Char(' ').Then(_ => WordParser).Many()
        select leading + " " + String.Join(" ", rest);

Can probably still clean it up a bit, but the concept is there.

Is this different from [`" I want to return up to this point ".Trim()`](https://learn.microsoft.com/en-us/dotnet/api/system.string.trim?view=netframework-4.8)? — grooveplex, May 13 '19 at 11:49
@grooveplex the OP is asking how to create a parser using a parser combinator. Not how to trim a string. Handling whitespace is the first step in almost any parser. Besides, `Trim()` generates temporary strings which can eradicate performance when parsing a lot of text or using a complex grammar — Panagiotis Kanavos, May 13 '19 at 11:50
@PanagiotisKanavos Correct,, Basically the premise is match any letter multiple times with only single spaces allowed in between untill more than one space is detected. — Heinrich Walkenshaw, May 13 '19 at 11:56
Seems your question is duplicate with https://stackoverflow.com/questions/206717/how-do-i-replace-multiple-spaces-with-a-single-space-in-c — Vadim Alekseevsky, May 13 '19 at 12:56
@VadimAlekseevsky not at all. The OP asked a question about *Sprache*, a parser combinator library. The linked question has nothing to do with it. — Panagiotis Kanavos, May 13 '19 at 12:57
@HeinrichWalkenshaw what have you tried? Post your attempts, it may be the only way to stop people from assuming you ask about `String.Trim`. Also check [this article](http://www.magnuslindhe.com/2014/09/parsing-whitespace-and-new-lines-with-sprache/). Have you tried using `Token()` or `WhiteSpace` ? Using parsers that match leading or trailing whitespaces it common, eg [like the leading/trailing](https://github.com/sprache/Sprache#usage) parsers in the library's example — Panagiotis Kanavos, May 13 '19 at 13:01
@HeinrichWalkenshaw the parser you want insn't all that different from a CSV parser, where the separator is a space instead of a comma. You could check the [record parser](https://github.com/sprache/Sprache/blob/master/test/Sprache.Tests/Scenarios/CsvTests.cs#L47) from the CSV example. — Panagiotis Kanavos, May 13 '19 at 13:08
Thank you @PanagiotisKanavos, the trick seems to be to use .Then(). Will update the question. — Heinrich Walkenshaw, May 13 '19 at 13:29

score 3 · Answer 1 · answered May 13 '19 at 13:34

Thanks to @PanagiotisKanavos, the trick would be to use the .Then() operator. The following works:

public static Parser<string> WordParser =
    Parse.Letter.Many().Text().Token();

public static Parser<string> PhraseParser =
    from leading in Parse.LetterOrDigit.Many().Text()
    from rest in Parse.Char(' ').Then(_ => WordParser).Many()
    select leading + " " + String.Join(" ", rest);

Parsing a phrase with Sprache(Words seperated by spaces)

1 Answers1