Re: Axis_group_engraver + Scheme engraver for staff combining (issue 576540043 by address@hidden)

[thomasmorley65]

On 2019/03/18 22:26:15, dak wrote:
> mailto:address@hidden writes:

> > On 2019/03/17 01:45:49, saul.james.tobin wrote:
> >
> >> Re cdr is not a predicate — the list being processed here is
composed
> > of
> >> pairs, the cdr of which is ##t or ##f.
> >
> > The description is still confusing. Do you mean something like
below?
> >
> > #(define (divide-true-cdr ls)
> > "Split @var{ls} into those elements which do and don't have a tail
of
> > value
> > @code{#t}"
> >  (call-with-values
> >    (lambda () (partition (lambda (x) (eq? #t (cdr x))) ls))
> >    (lambda (a b) (list a b))))
> >
> > #(display (divide-true-cdr '((1 . #f) (2 . #t) (3 . #f))))
> >
> > => (((2 . #t)) ((1 . #f) (3 . #f)))

> The silence of the lambdas (almost):

> #(define (divide-true-cdr ls)
>    "Split @var{ls} into those elements which don't and do have a tail
of
>    value @code{#tf}"
>    (call-with-values (lambda () (partition cdr ls)) list))

> #(display (divide-true-cdr '((1 . #f) (2 . #t) (3 . #f))))
> => (((2 . #t)) ((1 . #f) (3 . #f)))

> --
> David Kastrup


Aargh, replacing `(lambda (a b) (list a b))´ with `list´ is ofcourse far
better.
I considered to use simple `cdr´ instead of `(lambda (x) (eq? #t (cdr
x)))´, though one would need to ensure the `cdr´ is always a boolean,
otherwise our procedures would return differently sometimes.  Saul said
so, though I thought better be paranoid than sorry ...


https://codereview.appspot.com/576540043/