In your second example they both are RIGHT. When calling parse(@_) recursively the new instance of parse creates its own @_. When you shift @_ in the recursive call you are NOT shifting the @_ of the parent caller. Try this: 
#!/usr/bin/perl
  use strict;
  my $call = 0;
  parse(1,0);
  
  sub parse {
    my $curcall=++$call;
    print "My parameters: @_\n";
    while (defined(my $number = shift)) {
      print "Number: $number from call $curcall\n";
      parse(@_) if $number;
    }
  }

OUTPUT:

My parameters: 1 0
Number: 1 from call 1
My parameters: 0
Number: 0 from call 2
Number: 0 from call 1

[download]
Notice how the second printing of '0' is from call 1 and not call 2?

In reply to Re: Bug in Devel::DProf? by fletcher_the_dog
in thread Bug in Devel::DProf? by Ovid

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