Given an expression that specifies a hash element or array element, returns true if the
specified element in the hash or array has ever been initialized, even if the
corresponding value is undefined. The element is not autovivified if it doesn't exist.
print "Exists\n" if exists $hash{$key};
print "Defined\n" if defined $hash{$key};
print "True\n" if $hash{$key};
print "Exists\n" if exists $array[$index];
print "Defined\n" if defined $array[$index];
print "True\n" if $array[$index];
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A hash or array element can be true only if it's defined, and defined if it exists, but
the reverse doesn't necessarily hold true.
Given an expression that specifies the name of a subroutine, returns true if the
specified subroutine has ever been declared, even if it is undefined. Mentioning a
subroutine name for exists or defined does not count as declaring it. Note that a
subroutine which does not exist may still be callable: its package may have an AUTOLOAD
method that makes it spring into existence the first time that it is called -- see perlsub.
print "Exists\n" if exists &subroutine;
print "Defined\n" if defined &subroutine;
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Note that the EXPR can be arbitrarily complicated as long as the final operation is a
hash or array key lookup or subroutine name:
if (exists $ref->{A}->{B}->{$key}) { }
if (exists $hash{A}{B}{$key}) { }
if (exists $ref->{A}->{B}->[$ix]) { }
if (exists $hash{A}{B}[$ix]) { }
if (exists &{$ref->{A}{B}{$key}}) { }
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Although the deepest nested array or hash will not spring into existence just because
its existence was tested, any intervening ones will. Thus $ref->{"A"}
and $ref->{"A"}->{"B"} will spring into existence
due to the existence test for the $key element above. This happens anywhere the arrow
operator is used, including even:
undef $ref;
if (exists $ref->{"Some key"}) { }
print $ref; # prints HASH(0x80d3d5c)
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This surprising autovivification in what does not at first--or even second--glance
appear to be an lvalue context may be fixed in a future release.
See perlref/"Pseudo-hashes:
Using an array as a hash" for specifics on how exists() acts when used on a
pseudo-hash.
Use of a subroutine call, rather than a subroutine name, as an argument to exists() is
an error.
exists ⊂ # OK
exists &sub(); # Error
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