"Roy Chen" <[EMAIL PROTECTED]> wrote > def f(a, L=[]): > L.append(a) > return L
This creates a single list object L used for all calls to f() that don;t provide an explicit L. Thus each call appends to the same object. > > def g(a, L=None): > if L == None: > L = [] > L.append(a) > return L This creates a new L on each invocation of g() where L is None or not supplied. > print f(1) > print f(2) > print f(3) > print g(1) > print g(2) > print g(3) > > The output is: > [1] > [1, 2] > [1, 2, 3] The same default object returned each time > [1] > [2] > [3] 3 new objects returned. > I understand that default arguments are evaluated once only, at the > point of function definition, but I can't wrap my head around how > the > function g() doesn't do the same thing as the function f(). Because it creates a new object at execution not at definition. HTH, -- Alan Gauld Author of the Learn to Program web site http://www.freenetpages.co.uk/hp/alan.gauld _______________________________________________ Tutor maillist - Tutor@python.org http://mail.python.org/mailman/listinfo/tutor
