On Friday, July 10, 2015 at 8:04:36 AM UTC-4, candide wrote: > Le vendredi 10 juillet 2015 04:02:56 UTC+2, Chris Angelico a écrit : > > > > > I'm not sure what contradiction you're referring to, here. The > > evaluation that you're pointing out says, as Terry showed via the > > disassembly, that Python's first action is to look up the name 't' and > > grab a reference to whatever object it points to. > > > But in order to perform an operation, the interpreter has to evaluate the > operands and "evaluating" is not "grabbing a reference to". > > > The execution of > > t.sort() has to happen before the multiplication, because of the > > parentheses. > > > > > > Official docs explains what evaluation is : > > When the name is bound to an object, evaluation of the atom yields that > object. > > So, since the Python interpreter is performing evaluation from left to right, > the first operand of the expression : > > t*(1+int(bool(t.sort()))) > > evaluates to [2020, 42, 2015]. Next, the second operatand evaluates to the > integer 1. So I was expecting the result to be a shallow copy of the first > list [2020, 42, 2015] (the value of t before side effect produced by the sort > method). On the contrary, the final result takes into in account the side > effect and it is as if the first operand has been evaluated twice before > execution of the multiplication operation.
The first operand is t. Evaluating t does not make a copy of t, it is simply a reference to t. If t is later modified (by the sort method), the modified data will be seen when t is used in the multiplication. Python never implicitly copies lists (or any other data structure). This explains more about the mechanics of names and values: http://bit.ly/pynames1 --Ned. -- https://mail.python.org/mailman/listinfo/python-list
