Op 2005-11-03, Stefan Arentz schreef <[EMAIL PROTECTED]>: > Antoon Pardon <[EMAIL PROTECTED]> writes: > > ... > >> Fine, we have the code: >> >> b.a += 2 >> >> We found the class variable, because there is no instance variable, >> then why is the class variable not incremented by two now? > > Because it really is executed as: > > b.a = b.a + 2
That is an explanation, not a reason. > 1. get 't'b.a and store it in a temporary 't' (found the instance) > 2. add 2 to 't' > 3. store 't' in 'b.a' > > The last operation stores it into an instance variable. [ I think you mean '(found the class variable)' in line 1 ] All you are doing here is explain how the current implemantation treats this. You are not giving arguments for why the result of this implementation should be considered sane behaviour. >> > Remember, Python is a dynamic language. >> >> So? Python being a dynamic language doesn't prevent the following to fail: >> >> >>> a=1 >> >>> def f(): >> ... a += 2 >> ... >> >>> f() >> Traceback (most recent call last): >> File "<stdin>", line 1, in ? >> File "<stdin>", line 2, in f >> UnboundLocalError: local variable 'a' referenced before assignment > > See the 'global' keyword. You missed the point. If python being a dynamic language would be the answer to the interaction between instance and class variable, why then is the interaction between local and global variable different, why wouldn't the f code be executed as follows 1. get a and store in in a temporary 't' (found the global) 2. add 2 to 't' 3. store 't' in 'a' The last operation storing it in f's local namespace. -- Antoon Pardon -- http://mail.python.org/mailman/listinfo/python-list
