Re^6: scanning hash

To illustrate the point:

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

Comment on Re^6: scanning hash Download Code

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Re^7: scanning hash by ysth (Canon) on Aug 02, 2004 at 20:20 UTC
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 $y); # prints 1 for me` [download] or `$x = 0; $y = "0";` [download] Are they equal? Yes and no.	[reply] [d/l] [select]
Re^8: scanning hash by Aristotle (Chancellor) on Aug 02, 2004 at 20:37 UTC
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.	[reply]