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Why does printf() do this to big numbers?

by Cody Fendant (Friar)
on Mar 17, 2010 at 06:11 UTC ( #829093=perlquestion: print w/replies, xml ) Need Help??
Cody Fendant has asked for the wisdom of the Perl Monks concerning the following question:

printf( "%s\n", 2810337464 ); ## prints 2810337464 printf( "%d\n", 2810337464 ); ## prints -1484629832

Using ActiveState Perl 5.10 on Windows XP.

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Re: Why does printf() do this to big numbers?
by Anonymous Monk on Mar 17, 2010 at 06:13 UTC
Re: Why does printf() do this to big numbers?
by fullermd (Priest) on Mar 17, 2010 at 06:31 UTC

    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.

      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
      $ 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 ...

        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.

      $ 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';
Re: Why does printf() do this to big numbers?
by ikegami (Pope) on Mar 17, 2010 at 16:06 UTC

    Max 32-bit signed integer: 2,147,483,647
    Max 32-bit unsigned integer: 4,294,967,295

    You want to print 2,810,337,464, which lands between the two. You have a 32-bit build of Perl and you told Perl to print a signed integer. It started by converting the input to an integer if it wasn't already, then it treated the result as a signed integer. The result is 2810337464 - 2**32

    Ask for an unsigned integer or a floating point number:

    $ perl -e'printf( "%d\n", 2810337464 );' -1484629832 $ perl -e'printf( "%u\n", 2810337464 );' 2810337464 $ perl -e'printf( "%.0f\n", 2810337464 );' 2810337464
Re: Why does printf() do this to big numbers?
by ssandv (Hermit) on Mar 17, 2010 at 18:38 UTC
    This might be of some interest.

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