Re^2: Why does printf() do this to big numbers?

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
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$ 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
                        ...
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Re^3: Why does printf() do this to big numbers? by fullermd (Vicar) 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]