Re: A problem with range objects and floating point numbers

[s.jo]

Well, I don't think that way on the printf format:

octave:18> sprintf('%.43g',0.1)
ans = 0.1000000000000000055511151231257827021181583

Do you think that the above ans is rounded around 12th position?
Conversely, the decimation at 12th position is the way how Matlab shows the
number in display
only when the explicit precision number is not specified such as '%12f' or
'%.43g'.
Even though Matlab shows that way on display, any raw bits are NOT
discarded.
You can check it by using num2hex command in matlab. 

Anyhow, it is fun to check the detail behaviors:
Here are the test result up-to test 8:

% generate ranges : 
%  1. range object, 2. matrix object, 3. linspace, 4. perfect decimation
N=1000;
range_test1=(-N:0.1:N);
range_test2=[-N:0.1:N];
range_test3=linspace(-N,N,20*N+1);
range_test4=(str2num(sprintf('%.12f ',-N:0.1:N)));
range_test5=(-N/0.1:1:N/0.1)*0.1; 
range_test6=[-N/0.1:1:N/0.1]*0.1; 
% same as range_test6
NN=20*N+1; 
range_test7=zeros(1,NN); 
for i=(1:1:NN) 
   range_test7(i)=(i-1-N/0.1)*0.1; 
endfor 
% the linear spacing (interpolation) algorithm
range_test8=zeros(1,NN); A=-N; B=N; 
for i=(1:1:NN) 
   range_test8(i)=(A*(NN-i) + B*(i-1))/(NN-1); 
endfor 

+ sum (abs (sin (range_test1 * pi)) < eps (range_test1 * pi))
ans =       1055
+ sum (abs (sin (range_test2 * pi)) < eps (range_test2 * pi))
ans =       1168
+ sum (abs (sin (range_test3 * pi)) < eps (range_test3 * pi))
ans =       2001
+ sum (abs (sin (range_test4 * pi)) < eps (range_test4 * pi))
ans =       2001
+ sum (abs (sin (range_test5 * pi)) < eps (range_test5 * pi))
ans =       1055
+ sum (abs (sin (range_test6 * pi)) < eps (range_test6 * pi))
ans =       2001
+ sum (abs (sin (range_test7 * pi)) < eps (range_test7 * pi))
ans =       2001
+ sum (abs (sin (range_test8 * pi)) < eps (range_test8 * pi))
ans =       2001

The results are as you expected.

-- jo



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