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?
| [reply] |
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
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] |