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pglenski has asked for the wisdom of the Perl Monks concerning the following question:
Back in 04/27/01 #76296 dvergin posted a solution to finding unique occurrences in a hash. It works beautifully but I can't figure how it works.
my @ary = qw(a b c d d e b d);
my %hsh;
undef @hsh{@ary};
my @ary2 = keys %hsh;
print "@ary2\n";
It's the line undef @hsh{@ary} that I can't figure out. I need baby step explanations, I guess. Isn't @hsh and %hsh two completely different things?
Thanks in advance. Pls refer to the what page # in the Camel book if possible.
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Re: hash slice
by Roy Johnson (Monsignor) on Mar 10, 2005 at 15:29 UTC
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Page 37 gives an example of a hash slice. The important concept to note is that the sigil (the symbol in front of a variable name, like $, @, or %) connotes what kind of value is being retrieved, not what kind of variable is being accessed.
The shape of the braces indicate that @hsh{@ary} is a slice of a hash, rather than a slice of an array, which would be @hsh[@ary].
Applying undef to a hash slice relies on undocumented behavior, but the intent is to get all those keys to exist in the hash. A better technique is
@hsh{@ary} = ();
Caution: Contents may have been coded under pressure.
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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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Being unknowing, is not the same as being stupid.
Expressing a contrary opinion, whether to the individual or the group, is more often a sign of deeper thought than of cantankerous belligerence.
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Basically, undef is a keyword that takes an array.
Yes, but a slice is not an array, and undef is explicitly not a listop.
undef
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.
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Re: hash slice
by Mugatu (Monk) on Mar 10, 2005 at 19:28 UTC
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I apologize if you already got this from the other answers, but you requested baby steps, and I didn't see this particular baby step being explained. :-)
Isn't @hsh and %hsh two completely different things?
Yes, @hsh and %hsh by themselves are two different things, except you never had @hsh by itself in your code. You had @hsh{...}, which is, as you surmised, the hash slice syntax. It has an @ in front because it's accessing a list of values, but the variable that it's accessing from is %hsh.
And when I say a list of values, that's exactly what I mean. Hash slice syntax is a convenient way to get access to more than one hash value at a time. It is equivalent to a list of single hash accesses:
# given a hash:
my %hash = (one => 1, two => 2);
# the slice:
print @hash{"one", "two"};
# is equivalent to:
print $hash{"one"}, $hash{"two"};
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