On Monday 30 November 2009, Hugo Arts wrote: > Consider this python session: > >>> x = 0 > >>> def f(): > > x = x + 1 > > >>> f() > > Traceback (most recent call last): > File "<pyshell#27>", line 1, in <module> > f() > File "<pyshell#26>", line 2, in f > x = x + 1 > UnboundLocalError: local variable 'x' referenced before assignment Ah... what a pity I didn't try this. I used similar code in my response to Spir and thought it would be somehow connected to closures. Sending nonsense statements to the list again...
> > The python docs offers some insight: > > The execution of a function introduces a new symbol table used for > the local variables of the function. More precisely, all variable > assignments in a function store the value in the local symbol > table; whereas variable references first look in the local symbol > table, then in the local symbol tables of enclosing functions, then > in the global symbol table, and finally in the table of built-in > names. Thus, global variables cannot be directly assigned a value > within a function (unless named in a global statement), although > they may be referenced. > > ( from > http://docs.python.org/tutorial/controlflow.html#defining-functions > ) > > In short, the problem is that writing "x =" will create a new > (unbound) local name for x, hiding the global one. The following > reference to x will find the local unbound variable and start > complaining. I find that behavior quite counterintuitive. I expect: 1. 'x' is looked up, global variable 'x' is found; 2. the addition is performed; 3. a local name 'x' is created and bound to the result of the addition. Producing the error is not insane. Because referencing a global variable and shadowing it in the same statement is bad style. Eike. _______________________________________________ Tutor maillist - Tutor@python.org To unsubscribe or change subscription options: http://mail.python.org/mailman/listinfo/tutor
