Re^2: Confused by reference sample code

Many thanks - I must admit that the difference between 'my' inside a loop and outside in this context had elluded me (the distinction between 'new variable' and 'same variable' is somewhat blurred in this context).

Question: if the memory allocated to the hash had not been passed by ref. to the array, would perl have released it's memory?

Tom Melly, tom@tomandlu.co.uk

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Re^3: Confused by reference sample code by ikegami (Patriarch) on Mar 28, 2006 at 22:32 UTC
if the memory allocated to the hash had not been passed by ref. to the array, would perl have released it's memory? Yes. Perl uses reference counting to determine if a variable is still in use. As soon as there are no references to a variable, the memory is released back to Perl (but not necessarilly to the operating system). For example, `{ # Creates a scalar with value 'val1'. my $var1 = 'val1'; # 1 reference to val1. } # $var1 goes out of scope. # 0 references to val1. # No more references to 'val1', so it is released. my $var2; { # Creates a scalar with value 'val1'. my $var3 = 'val2'; # 1 reference to val2. $var2 = \$var3; # 2 references to val2. } # $var3 goes out of scope. # 1 reference to val2. $var2 = 'val3' # 0 references to val2. 1 reference to val3. # No more references to 'val2', so it is released.` [download]	[reply] [d/l]

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Re^3: Confused by reference sample code
by ikegami (Patriarch) on Mar 28, 2006 at 22:32 UTC

if the memory allocated to the hash had not been passed by ref. to the array, would perl have released it's memory?

Yes. Perl uses reference counting to determine if a variable is still in use. As soon as there are no references to a variable, the memory is released back to Perl (but not necessarilly to the operating system). For example,

{
   # Creates a scalar with value 'val1'.
   my $var1 = 'val1';  # 1 reference to val1.
}

# $var1 goes out of scope.
# 0 references to val1.
# No more references to 'val1', so it is released.

my $var2;
{
   # Creates a scalar with value 'val1'.
   my $var3 = 'val2';  # 1 reference to val2.
   $var2 = \$var3;     # 2 references to val2.
}

# $var3 goes out of scope.
# 1 reference to val2.

$var2 = 'val3'  # 0 references to val2. 1 reference to val3.

# No more references to 'val2', so it is released.
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