Why does printf() do this to big numbers?

Cody Fendant has asked for the wisdom of the Perl Monks concerning the following question:

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Re: Why does printf() do this to big numbers? by Anonymous Monk on Mar 17, 2010 at 06:13 UTC
Because that is what printf does , and that is how computer numbers work :) No, really :) What Every Computer Scientist Should Know About Floating-Point Arithmetic	[reply]
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.	[reply] [d/l]
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]
Re^3: Why does printf() do this to big numbers? by fullermd (Priest) on Mar 18, 2010 at 00:25 UTC
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]
Re: Why does printf() do this to big numbers? by ikegami (Patriarch) 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` [download]	[reply] [d/l] [select]
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.	[reply]


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