Now you're modifying an array, swapping one value (a list) for another (another list). Calling the array @list doesn't change the fact it's an array. The code below doesn't modify '3' either - it just replaces the value of a variable. 
   my $var = 3;
   $var++;
   say $var;  # Prints 4.

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In reply to Re^8: Perl vs C by JavaFan
in thread Perl vs C by santhosh_89

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