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Re^5: scanning hash

by ysth (Canon)
on Aug 02, 2004 at 19:50 UTC ( #379404=note: print w/ replies, xml ) Need Help??


in reply to Re^4: scanning hash
in thread scanning hash

(puzzled) If you really want...I'll start us off.

You can't determine the equality of two values from the equality of their string representations.


Comment on Re^5: scanning hash
Re^6: scanning hash
by Aristotle (Chancellor) on Aug 02, 2004 at 20:13 UTC

    To illustrate the point:

    $ perl -le '$a = $b = "foo"; print \$a eq \$b ? "true" : "false"' false

    The strings are identical, but they're two different strings. The string representations of references to them differ. Are they equal or not? This is a question of semantics that you can't answer without additional context.

    Indeed, if we just accept the string representation as the universal measure of equality, then ambrus is right: the empty string and undef are equal, and thus all of my arguing has been moot. However, experience tells me it is a prudent bet to assume that in most situations, undef needs to be treated differently from an empty string.

    Makeshifts last the longest.

      I was thinking of something simpler, such as:
      $x = 1.23456; $y = 1.234560000000001; print 0+($x == $y); # prints 0 for me print 0+($x eq $y); # prints 1 for me
      or
      $x = 0; $y = "0";
      Are they equal? Yes and no.

        That is a very similar class of problem, although not exactly the same.

        I was dealing with the question of whether to compare "by reference" or "by value". A typical situation where this is an issue is comparing objects. What is the criterion then, identity or equality? The answer is much less clear than with the strings I used to demonstrate the point, because objects manifest as references by definition.

        Makeshifts last the longest.

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