efloat example

>>> for p in (3,6,9):
...     EFloat.precision = p
...     for f in (math.pi, 12, 823, 1024, 1024**2, 1024**3):
...         print EFloat(f), 1/EFloat(f)
... 
   3.142e+00  318.310e-03
  12.000e+00   83.333e-03
 823.000e+00    1.215e-03
   1.024e+03  976.562e-06
   1.049e+06  953.674e-09
   1.074e+09  931.323e-12
   3.141593e+00  318.309886e-03
  12.000000e+00   83.333333e-03
 823.000000e+00    1.215067e-03
   1.024000e+03  976.562500e-06
   1.048576e+06  953.674316e-09
   1.073742e+09  931.322575e-12
   3.141592654e+00  318.309886184e-03
  12.000000000e+00   83.333333333e-03
 823.000000000e+00    1.215066829e-03
   1.024000000e+03  976.562500000e-06
   1.048576000e+06  953.674316406e-09
   1.073741824e+09  931.322574615e-12

Dear Keith, that is exactly how it should be! (I cross-checked with a HP calculator to make sure.)
Hopefully, somebody will pick this Python issue up now.