On Wed, 01 Apr 2009 19:13:33 +0200, TP wrote: > Hi everybody, > > Try the following python statements: > >>>> "%.40f" % 0.2222222222222222222222222222222 > '0.2222222222222222098864108374982606619596' >>>> float( 0.2222222222222222222222222222222) > 0.22222222222222221
Remove the leading quote from the first one, and you'll see that the two numbers look pretty similar: 0.2222222222222222098864108374982606619596 0.22222222222222221 By the way, calling float(0.2222...2) is redundant, because 0.222...2 is already a float. Calling float again just wastes CPU cycles, because the same object is returned again. >>> x = 0.2222222222222222222222222222222222 >>> x is float(x) # check for object identity (same memory address) True We can see that floats have more precision than they display by default: >>> x 0.22222222222222221 >>> x - 0.2222222222222222 == 0 # Sixteen of digit 2 True >>> x - 0.222222222222222 # Fifteen of digit 2 2.2204460492503131e-16 Notice that doing this reveals more significant digits than were apparent from just printing x. > My problem is the following: > * the precision "40" (for example) is given by the user, not by the > programmer. > * I want to use the string conversion facility with specifier "e", that > yields number is scientific format; so I cannot apply float() on the > result of "%.40e" % 0.2222222222222222222222222222222, I would lost the > scientific format. No, this is confused. The float you create is the exact same object whether you use scientific format or not. >>> a = 0.0123 >>> b = 1.23e-2 >>> a == b True >>> a 0.0123 >>> b 0.0123 *All* floats contain mantissa and an exponent, but in binary, not decimal: >>> math.frexp(a) (0.78720000000000001, -6) >>> 0.78720000000000001 * 2**-6 0.0123 > Is there any means to obtain the full C double in Python Floats in Python *are* C doubles. -- Steven -- http://mail.python.org/mailman/listinfo/python-list
