How to understand closure in a lambda?

Question

I want to make 5 buttons in a loop, and for each buttons bind a commend to print the index. In the following solution it always prints the same index.

My code like this:

for i in range(5):
    make_button = Tkinter.Button(frame, text ="make!", 
                                 command= lambda: makeId(i))

def makeId(i):
    print(i)

It always prints 5. How can I fix this?

Asked By: simotophs23

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Source

Answer 1

Resolution of variables in lambdas is done when lambda is executed. At this time, for all buttons i=5. To rectify this issue do as follows:

 make_button = Tkinter.Button(frame, text ="make!", 
                              command= lambda i=i: makeId(i))

This creates i as a local variable in a lambda. This local variable will hold correct value of i from the loop. the local variable can have any name, not necessarily i, e.g. command= lambda a=i: makeId(a)).

Answered By: Marcin

How to understand closure in a lambda?

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