Garrett Cooper gcoo...@freebsd.org wrote:
I was talking to someone today about this macro, and he noted that
the algorithm is incorrect -- it fails the base case with ((x) == 0 --
which makes sense because 2^(x) cannot equal 0 (mathematically
impossible, unless you consider the limit as x
On Thursday, October 14, 2010 11:49:23 pm Garrett Cooper wrote:
On Thu, Oct 14, 2010 at 6:37 AM, John Baldwin j...@freebsd.org wrote:
On Thursday, October 14, 2010 7:58:32 am Andriy Gapon wrote:
on 14/10/2010 00:30 Garrett Cooper said the following:
I was talking to someone today about
On Fri, Oct 15, 2010 at 5:51 AM, John Baldwin j...@freebsd.org wrote:
On Thursday, October 14, 2010 11:49:23 pm Garrett Cooper wrote:
On Thu, Oct 14, 2010 at 6:37 AM, John Baldwin j...@freebsd.org wrote:
On Thursday, October 14, 2010 7:58:32 am Andriy Gapon wrote:
on 14/10/2010 00:30 Garrett
We aren't dealing with mathematicians, but programmers.
I am attempting to reconcile this with Colin's rationale in terms of
congruence classes over rings Z/nZ. ;)
b.
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on 14/10/2010 00:30 Garrett Cooper said the following:
I was talking to someone today about this macro, and he noted that
the algorithm is incorrect -- it fails the base case with ((x) == 0 --
which makes sense because 2^(x) cannot equal 0 (mathematically
impossible, unless you consider
On Thursday, October 14, 2010 7:58:32 am Andriy Gapon wrote:
on 14/10/2010 00:30 Garrett Cooper said the following:
I was talking to someone today about this macro, and he noted that
the algorithm is incorrect -- it fails the base case with ((x) == 0 --
which makes sense because 2^(x)
On Thu, Oct 14, 2010 at 6:37 AM, John Baldwin j...@freebsd.org wrote:
On Thursday, October 14, 2010 7:58:32 am Andriy Gapon wrote:
on 14/10/2010 00:30 Garrett Cooper said the following:
I was talking to someone today about this macro, and he noted that
the algorithm is incorrect -- it
I was talking to someone today about this macro, and he noted that
the algorithm is incorrect -- it fails the base case with ((x) == 0 --
which makes sense because 2^(x) cannot equal 0 (mathematically
impossible, unless you consider the limit as x goes to negative
infinity as log (0) / log(2)
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