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Creating a new one. It isn't a closure, so: x = 3 def f(x): # note the name here x = 20 f(x) x # => 3 Similarly, using a different parameter name:

Creating a new one. It isn't a closure, so: x = 3 def f(x): # note the name here x = 20 f(x) x # => 3 Similarly, using a different parameter name: def g(y): x = y g(20) x # => 3 You have to explicitly mark the `x` in the function to be the global one to reference it: def h(y): global x x = y h(10) x # => 10 With a lambda it would be a closure: i = lambda y : x # throwing away y for fun i(3) # => 10 x = 20 i("hi") # => 20

fny 4y ago
`f` behaves just like closure. It can even be assigned to a variable. def f(x): x = 2 # <- this `x` is the same `x` as in the argument, it can be access via locals() internally You can even assign `f`. For example, `function = f`. Python is call by reference. Change my mind. `def f(x): x[0] = 1` will manipulate whatever object you pass to it.
- Jtsummers 4y ago
  If Python were fully call by reference then: def g(y): y = 3 x = 10 g(x) x # => ?? If it's 3 then it's pass by reference here, if it's 10 it's pass by reference. Which is it? Additionally, for something to be a closure it has to close over something (an environment). What does `f` or `g` close over? Note that they aren't changing any environment, they are "merely" functions, not closures. Python does have closures, but those aren't examples of them. And being able to assign a function to a variable does not make a closure, or do you think that C has closures because it has function pointers?
  - Jtsummers 4y ago
    Oops: If it's 3 then it's pass by reference here, if it's 10 it's pass by value. Which is it? (Don't hit reply when you're rushing out the door.)
- kazinator 4y ago
  > Python is call by reference. Change my mind. No need to change your entire mind; just an incorrect definition of call-by-reference. Passing a value which has reference semantics isn't call-by-reference. Your f receives x by value. That value is a reference into a boxed object. The x[0] = 1 does not change x; it changes the boxed object. >>> def f(x): ... x[0] = 1 ... >>> a = [ 0, 2, 3 ] >>> b = a >>> f(a) >>> a is b True The f function can do nothing to make "a is b" false. Under pass-by-reference, assigning to x would do that. Under pass-by-reference, we can pass a reference to a, such that a can be replaced, without affecting b. "pass-by {whatever}" refers to the semantics of the parameter and what it receives from the argument expression, and how, not the semantics of the argument object/value.
kazinator 4y ago
No, I mean this: def f(x): g = lambda : x x = 2 return g() f(42) What is does f(42) return? Holy fuck, I simply cannot just cut and paste the above into Python as-is, because of the "unexpected indent" which I added so that HN shows that as quote. It's complaining about an unexpected indent. Kill. Me. Now. Anywway: >>> def f(x): ... g = lambda : x ... x = 2 ... return g() ... >>> f(42) 2 So what it looks like is that there is only one x in the function, established by the parameter. This x is captured by the lambda, and is then reassigned the value 2. The lambda retrieves the 2. For completeness we show that if x is not assignmed, g accesses f's argument value: >>> def f(x): ... g = lambda : x ... return g() ... >>> f(42) 42
- deleted 4y ago
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fny 4y ago

`f` behaves just like closure. It can even be assigned to a variable. def f(x): x = 2 # <- this `x` is the same `x` as in the argument, it can be access via locals() internally You can even assign `f`. For example, `function = f`. Python is call by reference. Change my mind. `def f(x): x[0] = 1` will manipulate whatever object you pass to it.

Jtsummers 4y ago
If Python were fully call by reference then: def g(y): y = 3 x = 10 g(x) x # => ?? If it's 3 then it's pass by reference here, if it's 10 it's pass by reference. Which is it? Additionally, for something to be a closure it has to close over something (an environment). What does `f` or `g` close over? Note that they aren't changing any environment, they are "merely" functions, not closures. Python does have closures, but those aren't examples of them. And being able to assign a function to a variable does not make a closure, or do you think that C has closures because it has function pointers?
- Jtsummers 4y ago
  Oops: If it's 3 then it's pass by reference here, if it's 10 it's pass by value. Which is it? (Don't hit reply when you're rushing out the door.)
kazinator 4y ago
> Python is call by reference. Change my mind. No need to change your entire mind; just an incorrect definition of call-by-reference. Passing a value which has reference semantics isn't call-by-reference. Your f receives x by value. That value is a reference into a boxed object. The x[0] = 1 does not change x; it changes the boxed object. >>> def f(x): ... x[0] = 1 ... >>> a = [ 0, 2, 3 ] >>> b = a >>> f(a) >>> a is b True The f function can do nothing to make "a is b" false. Under pass-by-reference, assigning to x would do that. Under pass-by-reference, we can pass a reference to a, such that a can be replaced, without affecting b. "pass-by {whatever}" refers to the semantics of the parameter and what it receives from the argument expression, and how, not the semantics of the argument object/value.

kazinator 4y ago

No, I mean this: def f(x): g = lambda : x x = 2 return g() f(42) What is does f(42) return? Holy fuck, I simply cannot just cut and paste the above into Python as-is, because of the "unexpected indent" which I added so that HN shows that as quote. It's complaining about an unexpected indent. Kill. Me. Now. Anywway: >>> def f(x): ... g = lambda : x ... x = 2 ... return g() ... >>> f(42) 2 So what it looks like is that there is only one x in the function, established by the parameter. This x is captured by the lambda, and is then reassigned the value 2. The lambda retrieves the 2. For completeness we show that if x is not assignmed, g accesses f's argument value: >>> def f(x): ... g = lambda : x ... return g() ... >>> f(42) 42

deleted 4y ago
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