Internally, perl changes the data type of variables where necessary/possible in arithmetic to try to avoid loss of precision and/or overflow. So for example on a typical 64-bit system, repeated $x += 1 will cause the internal type to change from being initially a signed 64-bit int, to an unsigned 64-bit, and finally to a double - at which point it doesn't overflow, but may lose precision.

The exact details of what types are used at what points depends on the platform.

Dave.

In reply to Re: How does Perl handle arithmetic overflows? by dave_the_m
in thread How does Perl handle arithmetic overflows? by f77coder

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