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I don't quite get why the 'Substitutability' code snippet fails? Is it because the both vars 'point' to the same RAM address initialized with '3'? If that is th

• amithgeorge 12y ago
  Isn't because after the increment operation, x has the value 4. Thus even if y has the value 3 and equals 3, it doesn't equal x.
  • kelvin0 12y ago
    The code line which 'fails' is $y == 3 , I think we are both confused ...
• tel 12y ago
  Don't think about RAM. Instead, consider something like the mathematical expression x = 3 = y thus `x` and `y` are both just new names for the same identical value `3`. Since both variables reference the same thing then we ought to be able to substitute x for y whenever we like. Of course, in Perl `x` isn't really a variable but instead a name referencing a "slot" which can be mutated. That mutation breaks the reference and makes `x` hold more meaning than merely a "reference to 3". So then you cannot substitute it for `y` any longer. I think his example code snippet is a little obtuse, though.