Difference Subtracting Double Precision from Single Precision

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Can you explain the results for the lines below
a=99999999; b=single(a); b-a
1
a=20000001; b=single(a); b-a
-1

Respuesta aceptada

Matt Fig
Matt Fig el 19 de Sept. de 2012
Editada: Matt Fig el 20 de Sept. de 2012
Yes, have a look at
E = eps(b)
When you convert to single you are going to get the nearest single value to a within E. To see what you are dealing with, look at this:
a = (999999999-32):(999999999+34); b = single(a);
fprintf('%10.0f %10.0f \n',[a;double(b)])
Now when you do
c = b-a
MATLAB does the equivalent of this:
c2 = single(double(b)-a)
Also have a look at this:

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