in reply to Re^6: why does push not default to $_? (simple)
in thread why does push not default to $_?

sub push (\@;@) { CORE::push ( @{+shift}, @_ ? @_ : $_ ) }
already tried it, it's a trap forget it. If you do push (@a,@b) and @b is empty, then $_ will be pushed, and that's definetly not what you expect!

If there is any feature request, which might make sense on the wishlist for the next perl5 version, it's to add a prototype symbol for "list defaulting to $_". "_" is already a symbol for "scalar defaulting to $_" which must be placed as last symbol before ";".

So I suggest double underscore "__" shall mean "list defaulting to $_".

I remember slightly that in PBP one can find a somehow similar reasoning why not to use prototypes! *

If anybody knows a way to already implement push() this way, please try these testcases:

use subs 'push'; sub push (\@;@) { # print $_[0],$_[1]; CORE::push( @{+shift}, $_[0]? @_ :$_) ; } $\="\n"; my @a; $_="_"; print "--- one parameter"; push @a; print @a; print "--- long list of parameters"; @b=("x","y"); @a=(); push @a,@b; print @a; print "--- one element"; @a=(); push @a,"z"; print @a; print "--- one element list"; @a=(); @b=("z"); push @a,@b; print @a; print "--- empty list"; @a=@b=(); push @a,@b; print @a; __END__ --- one parameter _ --- long list of parameters xy --- one element z --- one element list z --- empty list _

Cheers Rolf

UPDATES: Chapter9 "Subroutines"/Subchapter "Prototypes" -> "Don't use prototypes" ... The examples show that they mostly confuse because they often force scalar context!