The reason for appending "E0" is that the value in numerical context is unchanged: "3E0" as an integer is 3 (just like "3E1" is 30; not because numification ignores everything from the "E" till the end). Your value would means zero.

And I don't think 0+$count is easier than $count =~ /E/. Nor faster, or much faster at least. Your /(\d+)$/ definitely isn't easy, nor likely very fast.

As you can guess from the generally unknown behaviour of what .. actually returns, neither requirements are very common. That's why I'm most happy with the status, like it is.


In reply to Re^3: .. operator and not including the condition of right operand by bart
in thread .. operator and not including the condition of right operand by eXile

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