Re^2: "2" | "8" = ":" and 2|8=10

Bitwise operators are the only place where there's a difference between strings and numbers.

No. It matters for post-increment as well.

BTW This difference does not fit into the philosophy of Perl, where you should not be able to distinguish between strings and numbers, so this is actually a language design flaw.

It's a deliberate decision, and well worth the offset. It allows you to use bitfields that aren't restricted to 32 (or 64) bits.

Comment on Re^2: "2" \| "8" = ":" and 2\|8=10

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Re^3: "2" \| "8" = ":" and 2\|8=10 by bart (Canon) on Nov 04, 2011 at 12:46 UTC
Bitwise operators are the only place where there's a difference between strings and numbers. No. It matters for post-increment as well. I think you're wrong there. Postincrement only works different for strings if they start with a letter. For strings like "199" it makes no difference: you just get 200. And other strings apparently are converted to a number first, as `$x = "2A"; $x++` produces 3. It's a deliberate decision, and well worth the offset. It allows you to use bitfields that aren't restricted to 32 (or 64) bits. Yes, I don't question the functionality, that is useful. But the typical Perl thing to do would have been to provide different operators for strings and for numbers. Just as with `=` vs `eq` and `+` vs `.`.	[reply] [d/l] [select]
Re^4: "2" \| "8" = ":" and 2\|8=10 by JavaFan (Canon) on Nov 04, 2011 at 12:59 UTC
I think you're wrong there. Postincrement only works different for strings if they start with a letter. For strings like "199" it makes no difference: you just get 200. And other strings apparently are converted to a number first, as $x = "2A"; $x++ produces 3. `$ perl -E '$x = $y = "A3"; $y + 0; $x++; $y++; say "$x $y"' A4 1 $` [download] The difference is, $y has an integer (as well as a string value), where $x doesn't. And hence, the post-increment value differs.	[reply] [d/l]