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The Sort Problem

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Luke Palmer
February 11, 2004 19:11
The Sort Problem
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I've been thinking about this problem which comes up in my code a lot:

    @sorted = sort { $^'bar').compute <=> $^'bar').compute }

Often the expressions on each side are even longer than that.  But one
thing remains:  both sides are exactly the same, substitute a $^b for a

I can see a couple less-than-desirable ways around this redundancy:

    @sorted = sort { infix:<=>( *($^a, $^b)ยป.foo('bar').compute ) }

Which doesn't work if .compute returns a list... not to mention its
horrible ugliness.  Another is to define a variant of sort (haven't had
much practice with A6 material recently; here we go!):

    multi sub sort (&block($) = { $_ } : *@data) {
        sort { block($^a) cmp block($^b) } @data;

    @sorted = sort { .foo('bar').compute } @unsorted;

Which has the disadvantage of forcing you to use C<cmp> and forcing an
ascending sort.

Any other ideas?  Is a more general solution necessary?


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