Re: creating a variable in an if statement

perl -wMstrict -e
"sub S1 { return int rand 2 ? ('one 1') : () }
 sub S2 { return int rand 2 ? ('one 2') : ('bad', 'news') }
 for (0 .. shift) {
   if (1 == (my @f = S1 || S2)) { print @f         }
   else                         { print 'not one'  }
   print qq(\n);
   }"  10
one 1
one 1
one 1
not one
one 2
not one
one 2
one 1
not one
one 1
not one
[download]

Comment on Re: creating a variable in an if statement Download Code

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Re^2: creating a variable in an if statement by shmem (Chancellor) on Jan 22, 2008 at 21:34 UTC
That's flawed since in `if (1 == (my @f = S1 \|\| S2)) {` [download] scalar context is forced upon the return of `S1`, but not on that of `S2`: `perl -le '@g = qw(a b c); @h = qw(d e); @f = @g \|\| @h; print @f' 3 perl -le '@g = qw( ); @h = qw(d e); @f = @g \|\| @h; print @f' de` [download] What if `S1` returns more than one element? --shmem _($_=" "x(1<<5)."?\n".q·/)Oo. G°\ / /\_¯/(q / ---------------------------- \__(m.====·.(_("always off the crowd"))."· ");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}	[reply] [d/l] [select]
Re^2: creating a variable in an if statement by ikegami (Patriarch) on Jan 22, 2008 at 21:36 UTC
In the OP's code, `sub S1 { return qw( a b ); } sub S2 { return qw( c ); }` [download] S2 would get called and the result would be `@f=qw(c)`. But with yours, S2 doesn't get called and the result is `@f=qw(b)`.	[reply] [d/l] [select]