Re: Re: How to portably determine integer limits?

There are reasons to worry, most notable with regards to the bitwise operands, &, |, ^, ~ on one hand, << and >> on the other. And % also has a limit on the range for which it works reliable — very system dependent, I've had a perl 5.005 for which modulo calculations with a number > 2**32 always returned zero, but on my newer port, I can go up to 2**52 or 2**53 (I don't recall exactly), the mantissa part of an IEEE double float, of 64 bits.

This meaningfulness does give a clue on how to efficiently test how large the range can be: I'm quite convinced the limit is always closely related to a power of 2 so one could shift a number to the left, until the result differs from the number times 2, in floating point calculation.

my $i = 0;
my $n = my $m = 1;
while($n == $m) {
    $n <<= 1;
    $m *= 2;
    $i++;
}
print "different for $m (2**$i)\n";
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I get:

different for 4294967296 (2**32)

What precisely this implies, I leave as an exercise for the reader. :)

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Re: Re: Re: How to portably determine integer limits? by exussum0 (Vicar) on Oct 30, 2003 at 05:26 UTC
Just in case you were not writing out your thoughts so the reader can learn... The limits are of a power of 2.. 'cause it's binary. For every extra bit you tack on, you are adding to a position representing 2^(N+1) position, after the most significant bit. Adding to a 1 bit number makes it 2 bits and a max of 3 (4-1). Adding another bit makes it a 3 bit number, and givig it a max of 7 (8-1), It's minus one just 'cause that's how binary works. 2^n would be 1000.... Same w/ char's (2^8 in ascii), unicode, (2^8 * 2 == 2^16).. With signed numbers, it's a bit different.. since your most significant bit, teh bit with teh largest 2^n position would be, is just a bit saying if the number is negative or not. But for more reference on how that works, look up 2's compliment on google. :) I 'm going to bed. Play that funky music white boy..	[reply]

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Re: Re: Re: How to portably determine integer limits?
by exussum0 (Vicar) on Oct 30, 2003 at 05:26 UTC

The limits are of a power of 2.. 'cause it's binary. For every extra bit you tack on, you are adding to a position representing 2^(N+1) position, after the most significant bit. Adding to a 1 bit number makes it 2 bits and a max of 3 (4-1). Adding another bit makes it a 3 bit number, and givig it a max of 7 (8-1), It's minus one just 'cause that's how binary works. 2^n would be 1000....

Same w/ char's (2^8 in ascii), unicode, (2^8 * 2 == 2^16)..

With signed numbers, it's a bit different.. since your most significant bit, teh bit with teh largest 2^n position would be, is just a bit saying if the number is negative or not. But for more reference on how that works, look up 2's compliment on google. :) I 'm going to bed.

Play that funky music white boy..

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