How do I hard-code variables in a dynamically created function in python?

Question

I have the following python3 code

class Test:
    pos = [0,0]
    actions = []
    def bar(self, target):
        for i in target:
            def _():
                print(i,end="")
            self.actions.append(_)
foo = Test()
foo.bar("abcd")
for i in foo.actions:
    i()

Which is meant to output: abcd

but instead it outputs: dddd

I'm pretty sure the function is using the value of i when executing (the last value i had) and not i's value the function _ is declared, which is what I want.

Each iteration of the loop doesn't create a new scope. So all the functions share the same i variable. — Barmar
– Barmar, Commented Aug 13, 2022 at 4:21
There is a second problem, the variable is at the class level you need to initialize actions in __init__ or it will be done at the class level, which is shared for all Test, not for foo which is a single instance of Test. — Cireo
– Cireo, Commented Aug 13, 2022 at 4:22

Mandera · Accepted Answer · 2022-08-13 05:00:50Z

2

The general solution to this is to store the value as a default parameter value, like this:

class Test:
    pos = [0,0]
    actions = []
    def bar(self, target):
        for i in target:
            def _(i=i):  # This is the only changed line
                print(i,end="")
            self.actions.append(_)
foo = Test()
foo.bar("abcd")
for i in foo.actions:
    i()

>>> abcd

edited Aug 13, 2022 at 5:00

answered Aug 13, 2022 at 4:22

Mandera

3,0873 gold badges23 silver badges30 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

throawaway account Over a year ago

Thanks, this seems to be the fastest solution

Daniel · Accepted Answer · 2022-08-13 04:29:22Z

2

Let's output some things:

class Test:
    pos = [0, 0]
    actions = []

    def bar(self, target):
        for i in target:
            print(f"i={i} id={id(i)}")

            def _():
                print(f"_i={i} _id={id(i)}")
                print(i, end="")

            self.actions.append(_)

Output:

i=a id=2590411675120
i=b id=2590411458416
i=c id=2590411377456
i=d id=2590411377200
_i=d _id=2590411377200
d_i=d _id=2590411377200
d_i=d _id=2590411377200
d_i=d _id=2590411377200

See, the i in def _ overrides every time for loop iterates and eventually last value is what you get.

How to solve this? Pass i as an argument:

from functools import partial


class Test:
    pos = [0, 0]
    actions = []

    def bar(self, target):
        for i in target:
            print(f"i={i} id={id(i)}")

            def _(i):
                print(f"_i={i} _id={id(i)}")
                print(i, end="")

            self.actions.append(partial(_, i))

Output:

i=a id=2618064721392
i=b id=2618064504688
i=c id=2618064423728
i=d id=2618064423472
_i=a _id=2618064721392
a_i=b _id=2618064504688
b_i=c _id=2618064423728

Let's remove print statements now:

from functools import partial


class Test:
    pos = [0, 0]
    actions = []

    def bar(self, target):
        for i in target:

            def _(i):
                print(i, end="")

            self.actions.append(partial(_, i))


foo = Test()
foo.bar("abcd")

for i in foo.actions:
    i()

# Output: abcd

answered Aug 13, 2022 at 4:29

Daniel

1871 silver badge3 bronze badges

1 Comment

med benzekri Over a year ago

actually i think you are bit mistaken, because the function _ isn't overridden and you can check this by inspecting the ids of the functions references in the actions list . saving reference of the function prevents the default behavior form happening thus a new function is defined (not sure why this happen)

med benzekri · Accepted Answer · 2022-08-13 15:01:29Z

the reason why you are getting the last value of i is that the function print takes reference to i and not literal value of it and since it's in a loop you will get the last value i had. a workaround to this problem would be as Mandera said to set function default parameter and this actually works because the value of i is stored in the function's attribute __defaults__, which is responsible of storing default parameters values.
So the final code would be :

class Test:
    pos = [0,0]
    actions = []
    def bar(self, target):
        for i in target:
            def foo(p=i):print(p,end="") 
            self.actions.append(foo)

           
if __name__=="__main__":
    foo = Test()
    foo.bar("abcd")
    
    for i in foo.actions:
      i()

Note
the function foo isn't overridden

Collectives™ on Stack Overflow

How do I hard-code variables in a dynamically created function in python?

3 Answers 3

1 Comment

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

1 Comment

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related