Re: Difference between Perl and Java for << operator?

The answer to your question is hidden in the documentation for Integer Arithmetic. You need use integer; if you want your result as a signed integer. (By default it is treated as an unsigned integer.) Note also, that your << 28 assumes that your perl is compiled for 32-bit integers.

Bill

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Re^2: Difference between Perl and Java for << operator? by syphilis (Archbishop) on Jul 31, 2021 at 10:25 UTC
Note also, that your << 28 assumes that your perl is compiled for 32-bit integers Well spotted - if you want to use the perl sub that the OP provided then: 1) you will need to be running under the `use integer` pragma; and 2) you will also need to be running a perl that was built with 32-bit integers. However, as has also been demonstrated, there are ways to obtain the desired result by doing a `0xFFFFFFFF << 28`, that will work on both 32-bit integer and 64-bit integer builds. Cheers, Rob	[reply] [d/l] [select]
Re^2: Difference between Perl and Java for << operator? by eyepopslikeamosquito (Archbishop) on Jul 31, 2021 at 07:33 UTC
Great find Bill! I missed this when desperately tuning my Game of Life simulation for extra speed (which uses way more bit operations than any Perl program I've ever written) -- back then, I just tried `use integer`, found it was a bit faster, all tests still passed... so just moved on. Should have carefully read the docs like you. :) Update: See also: Re^4: How to test for empty hash? (where `use integer` fixed a bug with the inchworm secret operator `~~$#foo`)	[reply] [d/l] [select]


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