defined is not meant to be used on an array. For starters, it would be impossible for an array to be undefined since undef is a scalar value.

As such, it appears that defined imposes a scalar context on its argument. An array name in scalar context is the number of elements in that array. Numbers are defined, so defined(@bar) always returns true.

Furthermore, my @bar = undef; doesn't create an empty array. It creates an array with one element in it (undef). my @bar = (); and my @bar; are equivalent and create empty arrays.

If you want to check if the array is empty, just use the array name in scalar context:

my @bar = undef; print("There are ", scalar(@bar), " elements in \@bar\n"); print("\@bar is empty\n") if !@bar; print("\@bar isn't empty\n") if @bar; my @foo; print("There are ", scalar(@foo), " elements in \@foo\n"); print("\@foo is empty\n") if !@foo; print("\@foo isn't empty\n") if @foo;
There are 1 elements in @bar @bar isn't empty There are 0 elements in @foo @foo is empty

You can check if an *array element* is defined, since array elements are scalars.

my @bar = (undef, 'a'); print(defined($bar[0])?1:0, "\n"); # 0 print(defined($bar[1])?1:0, "\n"); # 1 print(defined($bar[2])?1:0, "\n"); # 0 (doesn't exist)

Update: Clarified some details
Update: "Deleted" a paragraph in response to ysth's reply.


In reply to Re: undef==defined sometimes? (body question) by ikegami
in thread undef==defined sometimes? by perl-diddler

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