Michael Lazzaro writes:<snip>Forgive me; a very minor style & efficiency question... what would the canonical way to do this be, given what we know of Perl6?
# the hapless, inefficient version: return &result_of_some_big_long_calculation(...args...) if &result_of_some_big_long_calculation(...args...);
The obvious answers are this:
my bool $x = &result_of_some_big_long_calculation(...args...); return $x if $x;
That does something different, in that it has coerced the result into a
C<bool>[*0]. So after the first line C<$x> can only be 0 or 1[*1]. And
given that one of those states is false, the code becomes equivalent to:
However if you permit the function to return more than two different values (of which more than one are true) then it becomes a more interesting question.
My apologies, you are correct -- $x should _not_ be typed. It's the more interesting question I'm asking.
On Monday, March 31, 2003, at 12:30 PM, Paul wrote:
I started to suggest this myself, then realized that you might not want it to return at all if the value is false.
Yes, exactly:
sub foo(...args...) {
# We first attempt to get our return value the easy way.
# If successful (the resulting value is defined and true),
# just return that value. my $x = &baz(...args...);
return $x if $x; # Still here? OK, then... we've got a lot more work
# to do before we can return a reasonable value... }
I'm looking for a Perl6 way to say that oft-repeated, oft-chained two-line snippet up there without declaring the temporary variable. Using C<given> or C<when>, maybe?
MikeL
