Python copies references in lambda functions

Asked Feb 03 '21 at 20:10

Active Feb 03 '21 at 20:10

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I have encountered a problem in Python that I'd be happy if someone could help! I'm trying to execute the following code block:

def func(x,y):
    return x+y
    
if __name__=='__main__':
    
    func_list=[]
    
    for y in [1,2]:
        func_list.append(lambda x: func(x,y))
        
    print(func_list[0](0))
    print(func_list[1](0))

the output is

2
2

while what I expect is

1
2

I know (to some extent) that this happens because Python merges references when two variables are assigned equal. My question is, how to correct this code to obtain two lambda functions such that they behave expectedly as above?

asked Feb 03 '21 at 20:10

Mostafa Ayaz

Thank you both for the answers. Gonna check them! – Mostafa Ayaz Feb 03 '21 at 20:17
This isn't really about "copying references", this is how static scope works. `y` refers to *the variable in the scope where the function is defined*. If you `print(y)` you'll see its value is `2`. – juanpa.arrivillaga Feb 03 '21 at 20:18
1

Thanks @juanpa.arrivillaga, I see it. – Mostafa Ayaz Feb 03 '21 at 20:20
And it's very important to understand, this isn't a property of lambda expressions, this works the same way with a function definition statement – juanpa.arrivillaga Feb 03 '21 at 20:21

Python copies references in lambda functions

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