On Sun, 05 Dec 2004 12:59:40 +0200, Roie Kerstein <[EMAIL PROTECTED]> wrote:
>Hello! > >I want to compute a**b%c for very long numbers. >Computing a**b and then applying modulu is not practical, since it takes >ages. >There is a built-in optional parameter to the long.__pow__(a,b[,modulu]), >and it works well. >My question is: How can I write it is a form of an expression, like a**b? >I prefer not to fill my script with calls of the form long.__pow__(a,b,c). > You could subclass long to use a fixed modulus stored as a class variable, and make a factory function to return a class for a particular modulus, e.g., >>> def mmod(modulus): ... class M(long): ... def __pow__(self, p): ... return type(self)(long.__pow__(self, p, self.modulus)) ... def __repr__(self): return 'M(%s %%%s)'% (long.__repr__(self), self.modulus) ... M.modulus = modulus ... return M ... >>> M = mmod(7) >>> M <class '__main__.M'> >>> M(2**55) M(36028797018963968L %7) note that print will use the __str__ method of the underlying long >>> print M(2**55) 36028797018963968 >>> print M(2**55)**1 2 I made the power operation return the same type >>> M(10) M(10L %7) >>> M(10)**1 M(3L %7) >>> M(10)**2 M(2L %7) >>> M(10)**3 M(6L %7) >>> M(10)**4 M(4L %7) But if you use another operation, the result will be converted to the dominant type, since the only operation I overrode was __pow__. But you could override them all, by hand or automate the wrapping with a metaclass. >>> M(10)**4 *1 4L >>> M(10)**4 +0 4L >>> M(10)**4 +0.0 4.0 >>> M(10)**4 *1e1 40.0 You could also make M take an optional initialization parameter to override what's set with mmod. Or if you override all the time, not bother with mmod. Depends on what you need & how you are going to use the stuff. Regards, Bengt Richter -- http://mail.python.org/mailman/listinfo/python-list
