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But you've turned off the warning telling you that it doesn't looks_like_number. If you don't have warnings on, Perl assumes you know what you're doing and tries to do what you've asked it to with the (crummy) numberish data you've given it. If you have enabled warnings, it lets you know about it's misgivings but still goes ahead and does what you've asked.

I think the expectation mismatch is because Perl uses the (POSIX) C strtod(3) which by design (or at least by the manual page description :) follows this heuristic (from the start of the char* passed parse off as many digits followed by an optional "." and more digits followed by an optional exponent notation, and stop when you hit a not-digit-not-plusminus-not-exponent-marker character) rather than using if-looks_like_number-number-else-it's-zero. This is mentioned in perldata in passing, but if you're not familiar with the C API I'll grant that it's certainly open to causing confusion.

In reply to Re^3: is_numeric by Fletch
in thread is_numeric by tlm

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