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Re: A question on "##t"

From: Damian leGassick
Subject: Re: A question on "##t"
Date: Fri, 25 Jan 2008 00:36:23 +0000

well, that clears it up for me. thanks all

ralph, yes it was the scope of that first # that foxed me

and yes i assumed wrongly that the equals sign was the assignment operator

if anyone can boil this link down to a couple of sentences for the docs, i think there'd be a few people down the line who'd appreciate it



On 24 Jan 2008, at 22:49, Bertalan Fodor wrote:

Also if you used LilyPondTool you would see that it colors the scheme parts, so you would see where the scheme part end. See attached screenshot.


Damian leGassick írta:
actually, this confuses me too

if the # puts lilypond into scheme mode, does that mean that the equals-sign in #'merge-differently-headed = ##t is not scheme? if it is, then why not #'merge-differently-headed = #t ?


On 24 Jan 2008, at 16:06, Mats Bengtsson wrote:

Martin Seng Hin Yew wrote:
Bertalan Fodor (LilyPondTool) wrote:
Everything beginning with a # is a Scheme-language expression.
So this sets the property called 'merge-differently-headed to the value #t
#t is the expression meaning true in Scheme.

Hi Bertalan Fodor,

Okay...assume i knew the word "true" (means =yes or 1, right?), but ##t got double #, so what does the other # means?
He already told you :-) The first # tells LilyPond that "here comes a Scheme
expression", the "#t" which follows is the actual Scheme code.


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