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Re: Issue 1320: Rewriting bar-line::print

From: Marc Hohl
Subject: Re: Issue 1320: Rewriting bar-line::print
Date: Wed, 21 Mar 2012 10:50:53 +0100
User-agent: Mozilla/5.0 (X11; U; Linux x86_64; en-US; rv: Gecko/20120216 Thunderbird/3.1.19

Am 21.03.2012 03:20, schrieb David Kastrup:
Marc Hohl<address@hidden>  writes:

Is this a feasible approach? What am I currently doing wrong?

#(define bar-line-stencil-alist
   '(("|" . thin-stil)
     ("." . thick-stil)
     ("" . empty-stil)
           (thin-stil (bar-line::simple-bar-line grob hair extent rounded))
           (thick-stil (bar-line::simple-bar-line grob fatline extent rounded))
           (empty-stil (make-filled-box-stencil (cons 0 0) (cons 0 extent)))
           (stencil (assoc-get glyph bar-line-stencil-alist empty-stil))

That does not work.  stencil is set to a symbol, and that symbol is
never converted to a value.  When the function is being executed, the
_symbols_ thin-stil, stick-stil and empty-stil have long lost any
connection with the expressions that they were associated with while the
let-statement was being compiled.

You could write (local-eval stencil (the-environment)) to let the
lexical environment of the function compilation linger over long enough
to make this work, but you would earn a reward for the unnecessarily
most ugly and unstable code imaginable.

Instead, try using a case statement.  There is no need to calculate all
stencils and throwing most of them away.
That's what I originally did, but my idea was to provide an easy way to
add new stencils/bar lines without fiddling with bar-line::print.
But if this way is a no-go, I'll use case.

Thanks for your explanations!


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