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Re: maximum value of a scalar

by ikegami (Pope)
on Oct 21, 2008 at 21:58 UTC ( #718599=note: print w/ replies, xml ) Need Help??


in reply to maximum value of a scalar

For 32-bit version of Perl, the maximum integer (without using bigint or similar) that can be stored precisely is -253 on the negative side and 253 on the positive side.

>perl -e"printf qq{%.f\n}, 2**53+$_ for +1,0,-1" 9007199254740992 9007199254740992 9007199254740991 >perl -e"printf qq{%.f\n}, -2**53+$_ for +1,0,-1" -9007199254740991 -9007199254740992 -9007199254740992
build settingsMax positive integerMax negative integer
32-bit ints and double floats253 = 9_007_199_254_740_992-253 = -9_007_199_254_740_992
64-bit ints and double floats264-1 = 18_446_744_073_709_551_615-263 = 9_223_372_036_854_775_808
64-bit ints and quadruple floats2113 = 10_384_593_717_069_655_257_060_992_658_440_192-2113 = -10_384_593_717_069_655_257_060_992_658_440_192

Some operators (notably the bit operators) won't work with every number in that range.
Some C extensions won't work with every number in that range.


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Re^2: maximum value of a scalar
by Anonymous Monk on Mar 01, 2009 at 06:11 UTC
    I'm trying to compile for quadruple floats, and it's not obvious to me what the parameter is during configure. Suggestions? Thanks, Scott
      Sorry, none. I have no knowledge on the topic.
      I'm trying to compile for quadruple floats

      What is your compiler's C data type for these "quadruple floats" ? I had assumed that, if a C compiler could handle "quadruple floats", it would be handling them as "long doubles" - in which case building with -Duselongdouble might do the trick.

      Cheers,
      Rob
        I'm using gcc on linux. Not sure what the compiler's data type for "quadruple floats" is, though "long doubles" are 16 bytes accordint to ./Configure. It seems that (2^128) should be enough. I have long doubles turned on, and general 64 bit turned on, but I still max out at 2^64 instead of 2^113. Is there an additional gcc flag I can pass in that anyone knows of to get the 2^113 limit? Thanks! Scott

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