http://qs321.pair.com?node_id=829100


in reply to Why does printf() do this to big numbers?

Because that's what you asked for.

%d is asking for a signed int. On pretty much any platform out there, int is 32 bits. 2.8 billion is larger than can be represented in a signed int, ergo wraparound.

Replies are listed 'Best First'.
Re^2: Why does printf() do this to big numbers?
by almut (Canon) on Mar 17, 2010 at 09:26 UTC
    On pretty much any platform out there, int is 32 bits

    Not necessarily.  With a "64-bit perl", "%d" can handle ints larger than 32 bits just fine: 

    $ perl -e "printf qq(%d\n), 2810337464"
2810337464

    [download]
    $ perl -V
  ...
    intsize=4, longsize=8, ptrsize=8, doublesize=8, byteorder=12345678
    d_longlong=define, longlongsize=8, d_longdbl=define, longdblsize=1
+6
    ivtype='long', ivsize=8, nvtype='double', nvsize=8, Off_t='off_t',
+ lseeksize=8
  ...
  Compile-time options: MULTIPLICITY PERL_IMPLICIT_CONTEXT
                        PERL_MALLOC_WRAP THREADS_HAVE_PIDS USE_64_BIT_
+ALL
                        USE_64_BIT_INT USE_ITHREADS USE_LARGE_FILES
                        ...

    [download]
[reply]
[d/l]
[select]

      Ah, true. I guess my C-brain leaks through sometimes :) I guess that's probably the USE_64_BIT_INT.

      But at any rate, overflow is what the OP is seeing.

[reply]
Re^2: Why does printf() do this to big numbers?
by Anonymous Monk on Mar 18, 2010 at 05:01 UTC 
    $ perl -V:.+?size
d_chsize='define';
d_malloc_good_size='undef';
d_malloc_size='undef';
doublesize='8';
fpossize='8';
gidsize='4';
i16size='2';
i32size='4';
i64size='8';
i8size='1';
intsize='4';
ivsize='4';
longdblsize='12';
longlongsize='8';
longsize='4';
lseeksize='8';
nvsize='8';
ptrsize='4';
shortsize='2';
sig_size='27';
sizesize='4';
u16size='2';
u32size='4';
u64size='8';
u8size='1';
uidsize='4';
uvsize='4';

    [download]
[reply]
[d/l]

In Section Seekers of Perl Wisdom