Re: Okay, lets talk assemblers and Lua (was Re: A bit more on bits...)

[Sean Conner <sean@...>]

It was thus said that the Great Johann ''Myrkraverk'' Oskarsson once stated:
> 
> Ok, so I am trying out LPEG.  For somewhat rigid syntax, I came up with
> 
>    local p = { } -- the patterns, all in a table
>    p.space = lpeg.S( " \t\n\r" ) ^ 1 -- whitespace
>    p.label = lpeg.Cg( lpeg.R( "az" ) ^ 0, "lbl" ) * p.space
>    p.instruction = ( lpeg.R( "az" ) + lpeg.R( "AZ" ) ) ^ 1
>    p.operand = lpeg.Cg( lpeg.R( "09" ) ^ 1, "imm" )
>    p.line = lpeg.Ct(  p.label
>           * lpeg.Cg( p.instruction, "ins" ) * lpeg.S( " \t\n\r" ) ^ 0
>           * p.operand ) -- lbl, ins, op; in that order.
> 
> [I hope the spacing comes out OK; there was a discussion earlier about
> issues with Thunderbird, which I'm using.]

  It came out fine here.

> For lines that don't have a label, it creates a key with the empty
> string.  That is, the table will be something like
> 
>   {
>     imm = "35",
>     ins = "store",
>     lbl = ""
>   }
> 
> Is there a way for LPEG to return nil in that case instead?  

  Yes.  If you change p.label to read:

	p.label = (lpeg.Cg(lpeg.R("az")^1,'lbl') + lpeg.P(true))
	        * p.space

Then the following string " store 35" (note the leading space) will return:

	{
	  ins = "store",
	  imm = "35",
	}

A breakdown:

	p.label = (
		    -- check for at least one character 
		    -- that makes a label
		    lpeg.Cg(lpeg.R("az")^1,'lbl')

		    -- if there isn't a character, we still
	   	    -- want to succeed 
	 	    + lpeg.P(true) -- success, even though there's nothing
		)
	       * p.space -- followed by mandatory space characters.

You can also tighten up p.instruction to read:

	p.instruction = lpeg.R("az","AZ")^1

  -spc