in reply to Re: use64bitint in building Perl
in thread use64bitint in building Perl

uname -a is answered 'FreeBSD ... 6.3-RELEASE FreeBSD 6.3-RELEASE #2 ... root@fc:/usr/src/sys/i386/compile/VKERN i386'. This looks like a 32-bit kernel, right?

When I build without -Duse64bitint, Configure makes dependent directories named 'i386-freebsd'. With the switch, Configure says it's going to make directories named "i386-freebsd-64int". This seems like 64 bit ints are not the default. Or did I mistake your meaning?

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Re^3: use64bitint in building Perl
by zwon (Abbot) on Dec 29, 2008 at 23:13 UTC

    Yes, it's 32bit system.

    When I mentioned FreeBSD defaults, I mean that perl already installed on the system is built with -Duse64bitint. You can check this using

    perl -V | grep 64bitint

    If you build Perl yourself you should specify this option explicitly. If you build it from ports use PERL_64BITINT makefile option.

[reply]
[d/l]
      $ perl5.8.7 -V | grep 64bitint
    use64bitint=undef use64bitall=undef uselongdouble=undef

      [download]
      Looks like my IHP didn't built their perl in the way that your system's initial perl was built. I have always used a 5.8.8 that I built, but now for 5.8.9 I'm thinking about 64 bit ints and threading for the first time.

      Does anyone know if building perl with 64bitint means that ALL arithmetically-processed integers are stored in 8 bytes, or only those whose value requires more than 32 bits?
[reply]
[d/l]
        Does anyone know if building perl with 64bitint means that ALL arithmetically-processed integers are stored in 8 bytes

        The IV slot will always be 8 bytes - ie *all* integers will be stored in 8 bytes.

        Cheers,
        Rob
[reply]

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