[Numpy-discussion] assigning full precision values to longdouble scalars

Scott Ransom sransom at nrao.edu
Tue Feb 25 17:52:59 EST 2014

Hi All,

So I have a need to use longdouble numpy scalars in an application, and 
I need to be able to reliably set long-double precision values in them. 
  Currently I don't see an easy way to do that.  For example:

In [19]: numpy.longdouble("1.12345678901234567890")
Out[19]: 1.1234567890123456912

Note the loss of those last couple digits.

In [20]: numpy.float("1.12345678901234567890")
Out[20]: 1.1234567890123457

In [21]: numpy.longdouble("1.12345678901234567890") - 
Out[21]: 0.0

And so internally they are identical.

In this case, the string appears to be converted to a C double (i.e. 
numpy float) before being assigned to the numpy scalar.  And therefore 
it loses precision.

Is there a good way of setting longdouble values?  Is this a numpy bug?

I was considering using a tiny cython wrapper of strtold() to do a 
conversion from a string to a long double, but it seems like this is 
basically what should be happening internally in numpy in the above example!



Scott M. Ransom            Address:  NRAO
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