On Jul 13, 1:20 pm, Wayne Brehaut <[EMAIL PROTECTED]> wrote: > On Mon, 09 Jul 2007 23:51:25 -0700, "[EMAIL PROTECTED]" > > > > > > <[EMAIL PROTECTED]> wrote: > >On Jul 9, 11:42?pm, Paul McGuire <[EMAIL PROTECTED]> wrote: > >> On Jul 9, 11:21 pm, "Jim Langston" <[EMAIL PROTECTED]> wrote:> In Python > >> 2.5 on intel, the statement > >> > 2**2**2**2**2 > >> > evaluates to>>> 2**2**2**2**2 > > >> > 200352993040684646497907235156025575044782547556975141926501697371089405955 > >> > 63114 > >> > 530895061308809333481010382343429072631818229493821188126688695063647615470 > >> > 29165 > >> > 041871916351587966347219442930927982084309104855990570159318959639524863372 > >> > 36720 > > >> <snip> > > >> Exponentiation is right associative, so this is the same as: > > >> 2**(2**(2**(2**2))) > >> 2**2**2**4 > >> 2**2**16 > >> 2**65536 > > >> 2=10**0.3010, so 2**65536 is approx 10**19726 > > >> There are 19730 digits in your answer, > > >>>> import gmpy > >>>> n = 2**2**2**2**2 > >>>> gmpy.numdigits(n) > >19729 > > >Did you count the 'L'? > > numdigits(n)? > > What? 'L' is a digit in Python? I'm going back to Fortran! > > wwwayne > > > > >>so this seems to be at least in > >> the ball park. > > >> -- Paul- Hide quoted text - > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
'L' counts for 50, but only when you use Roman font. -- Paul -- http://mail.python.org/mailman/listinfo/python-list
