[BangPypers] Enclosing lexical context
Could some one explain to me this sentence, I read in an example online Python doesn't implement assignment of variables bound in an enclosing lexical context Example, a=[b] --Picachu ___ BangPypers mailing list BangPypers@python.org http://mail.python.org/mailman/listinfo/bangpypers
Re: [BangPypers] Enclosing lexical context
On Thu, Apr 15, 2010 at 10:36 PM, Picachu Nioto picachu.ni...@gmail.com wrote: Could some one explain to me this sentence, I read in an example online Python doesn't implement assignment of variables bound in an enclosing lexical context Example, a=[b] I'm not sure where you got this sentence means but Python's scoping is lexical but has 2 namespaces accessible from the current point of execution (locals and globals) which are arguably dynamic. Here's an example to show lexical scoping. http://pastebin.com/k2S5pvjZ You can see that it prints 1 which is the value of the free identifier foo lexically at the point of the print. As a counter example, here's an example in Emacs lisp which parallels the above but since it's dynamically scoped, the value printed is 2. http://pastebin.com/KX0JwC0u -- ~noufal http://nibrahim.net.in ___ BangPypers mailing list BangPypers@python.org http://mail.python.org/mailman/listinfo/bangpypers
Re: [BangPypers] Enclosing lexical context
Thanks for the prompt response Noufal! Could you explain, in what context is it mentioned here? http://pastebin.com/zLnL7yy8 http://pastebin.com/zLnL7yy8 On Thu, Apr 15, 2010 at 11:09 PM, Noufal Ibrahim nou...@gmail.com wrote: On Thu, Apr 15, 2010 at 10:36 PM, Picachu Nioto picachu.ni...@gmail.com wrote: Could some one explain to me this sentence, I read in an example online Python doesn't implement assignment of variables bound in an enclosing lexical context Example, a=[b] I'm not sure where you got this sentence means but Python's scoping is lexical but has 2 namespaces accessible from the current point of execution (locals and globals) which are arguably dynamic. Here's an example to show lexical scoping. http://pastebin.com/k2S5pvjZ You can see that it prints 1 which is the value of the free identifier foo lexically at the point of the print. As a counter example, here's an example in Emacs lisp which parallels the above but since it's dynamically scoped, the value printed is 2. http://pastebin.com/KX0JwC0u -- ~noufal http://nibrahim.net.in ___ BangPypers mailing list BangPypers@python.org http://mail.python.org/mailman/listinfo/bangpypers ___ BangPypers mailing list BangPypers@python.org http://mail.python.org/mailman/listinfo/bangpypers
Re: [BangPypers] Enclosing lexical context
On Thu, Apr 15, 2010 at 10:36 PM, Picachu Nioto picachu.ni...@gmail.comwrote: Could some one explain to me this sentence, I read in an example online Python doesn't implement assignment of variables bound in an enclosing lexical context a=[10] def f(x): ... a[0]=x ... print a ... f(2) [2] print a [2] In this case, the outer a is accessed automatically since we are using indices and there is no local list a, Python finds the scope from the global scope and assigns correctly. Example, a=[b] Looking at your code above, perhaps the 2nd explanation makes it clear. --Picachu ___ BangPypers mailing list BangPypers@python.org http://mail.python.org/mailman/listinfo/bangpypers -- --Anand ___ BangPypers mailing list BangPypers@python.org http://mail.python.org/mailman/listinfo/bangpypers
Re: [BangPypers] Enclosing lexical context
My reply got deleted partly before I sent it. Here is the full one. It is not clear from your email what context means but mostly it is the fact that Python differentiates between local and global scope when variables are assigned. For example, a=10 def f(x): ... a=x ... print a ... f(20) 20 print a 10 The global a is still unmodified since the f() function changes only its local a. To fix this we need to prefix the local a with global. def f(x): ... global a ... a=x ... print a ... f(20) 20 print a 20 However, this can also be done using a container as below. a=[10] def f(x): ... a[0]=x ... print a ... f(2) [2] print a [2] In this case, the outer a is accessed automatically since we are using indices and there is no local list a, Python finds the scope from the global scope and assigns correctly. Looking at your code above, perhaps the 2nd explanation makes it clearer and seems closer to what you are expecting as answer. --Anand On Thu, Apr 15, 2010 at 11:19 PM, Anand Balachandran Pillai abpil...@gmail.com wrote: On Thu, Apr 15, 2010 at 10:36 PM, Picachu Nioto picachu.ni...@gmail.comwrote: Could some one explain to me this sentence, I read in an example online Python doesn't implement assignment of variables bound in an enclosing lexical context a=[10] def f(x): ... a[0]=x ... print a ... f(2) [2] print a [2] In this case, the outer a is accessed automatically since we are using indices and there is no local list a, Python finds the scope from the global scope and assigns correctly. Example, a=[b] Looking at your code above, perhaps the 2nd explanation makes it clear. --Picachu ___ BangPypers mailing list BangPypers@python.org http://mail.python.org/mailman/listinfo/bangpypers -- --Anand -- --Anand ___ BangPypers mailing list BangPypers@python.org http://mail.python.org/mailman/listinfo/bangpypers
Re: [BangPypers] Enclosing lexical context
2010/4/15 Picachu Nioto picachu.ni...@gmail.com: Could some one explain to me this sentence, I read in an example online Python doesn't implement assignment of variables bound in an enclosing lexical context Example, a=[b] Consider the following example: a = 1 def f(): b = 2 def g(): c = 3 # this function can access all a, b and c variables. print a, b, c # c can be reassigned c = 42 # a can be reassigned only if it is declared as global, otherwise it is considered as local to this function global a a = 42 # b can't be reassigned because it is neither local nor global. It is in the enclosing lexical context SInce b can't be reassigned, the work-around is to modify the object instead of reassigning. However, python3.0 added a new nonlocal construct to enable that. With python 3, you should be able to say: nonlocal b b = 42 Anand ___ BangPypers mailing list BangPypers@python.org http://mail.python.org/mailman/listinfo/bangpypers
Re: [BangPypers] Enclosing lexical context
On Thu, Apr 15, 2010 at 23:33, Anand Chitipothu anandol...@gmail.com wrote: However, python3.0 added a new nonlocal construct to enable that. With python 3, you should be able to say: So there are globals, locals, and nonlocals. :) ___ BangPypers mailing list BangPypers@python.org http://mail.python.org/mailman/listinfo/bangpypers
