in reply to Re: hash slice
in thread hash slice

Applying undef to a hash slice relies on undocumented behavior, . . .

It's not undocumented behavior. It's just behavior that's documented in a different place, unrelated to hashslices. Basically, undef is a keyword that takes an array. @hsh{@ary} evaluates to an array. But, to do that, Perl needs to autovivify all the slots in the hash in order to return the values. Heck, anything that takes a list will do.

undef @hsh{@ary}; chomp @hsh{@ary}; chop @hsh{@ary}; map $_, @hsh{@ary}; 1 foreach @hsh{@ary};

Those will all do the trick. The undef is a no-op to provide something that requires evaluation of a list.

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Re^3: hash slice
by Roy Johnson (Monsignor) on Mar 10, 2005 at 15:54 UTC
    Basically, undef is a keyword that takes an array.
    Yes, but a slice is not an array, and undef is explicitly not a listop.


    Undefines the value of EXPR, which must be an lvalue. Use only on a scalar value, an array (using @ ), a hash (using % ), a subroutine (using & ), or a typeglob (using * ). ... Note that this is a unary operator, not a list operator.
    What happens when you undef a slice is that it is converted into a list, which is then evaluated in scalar context, which yields the last element of the slice as the scalar value. That value (and only that value) is then set to undef (so it's not a no-op; if you want a no-op use the \ operator, which will take a list). Contrast that with handing it an array proper. The entire array is then undefined.

    The conversion from hash slice to list to scalar is not documented, and should generate a warning. In fact, if you try to pass undef a list of array elements, you will get an error.

    there is no such thing as a list in scalar context

    Caution: Contents may have been coded under pressure.

      I thought that seeing it on my screen would make it easier to understand. I'm missing something here. I was kind of expecting the array to be unchanged ... but that's not what happened:

      $ perl -MData::Dumper -e '@x=0..10; undef @x[2..7]; print Dumper(\@x)' $VAR1 = [ 0, 1, 2, 3, 4, 5, 6, undef, 8, 9, 10 ];

      Obviously, if undef worked as is otherwise expected, all numbers from 2 through 7 would be undef. What surprised me here is that 7 still is undef'd. Something funky is happening.

      Not that I'd ever have this problem in production code - I don't ever have the need for this type of behaviour (undefing a slice)... and would have thought about using @x[2..7] = () first anyway.

      Update: Yeah, Roy Johnson explained it - I see that now. However, it's still a bit buried, and Roy Johnson's lengthier explanation below is much clearer. Thanks, Roy!

        It may have not been there when you started to respond, since I tend to revise and expand my answers after posting, but I did explain the behavior you see. I agree that it is not what most people expect.

        The slice is being converted internally into a list: ($x[2]..$x[7]), and then that list is being evaluated in a scalar context, which means that it is not a list, but an expression with multiple comma operators. And the comma operator yields its right-hand side.

        Interestingly, defined works the same way, so defined(@x[2..7]) is the same as defined($x[7]). So at least when you undef a hash slice, testing it for definedness works as expected, even though several of the values are defined.

        I note again, you cannot provide a list to undef or defined. You get a compiler error. For defined, you can say

        defined scalar($x[2],$x[3]) or warn "not defined!";
        and it does the same as it does for a slice. But you can't use scalar() with undef, because scalar doesn't yield lvalues.

        Caution: Contents may have been coded under pressure.