Why Does This Python Lambda Loop Print 6 Every Time? Mastering Late Binding

I stumbled upon a Python interview question: what does the following code output?

def testFun():
    temp = [lambda x : i*x for i in range(4)]
    return temp

for everyLambda in testFun():
    print (everyLambda(2))

My initial guess was the obvious sequence:

0
2
4
6

However, the actual answer was:

6
6
6
6

Running the code confirmed the latter result, which led me to realize the cause: Python’s closures use late binding, meaning the variable i is looked up when the inner function is called, not when it is defined. By the time the lambdas are invoked, the for loop has finished and i holds its final value 3, so each lambda computes 3 * 2 = 6.

To obtain the expected output 0, 2, 4, 6, we need to bind the current value of i at definition time. Below are four solutions.

Solution 1: Use a default argument to capture the current value.

def testFun():
    temp = [lambda x, i=i: i*x for i in range(4)]
    return temp

for everyLambda in testFun():
    print (everyLambda(2))

Solution 2: Employ functools.partial to fix the multiplier.

from functools import partial
from operator import mul

def testFun():
    return [partial(mul, i) for i in range(4)]

for everyLambda in testFun():
    print (everyLambda(2))

Solution 3: Write a generator expression that creates the lambdas with immediate binding.

def testFun():
    return (lambda x, i=i: i*x for i in range(4))

for everyLambda in testFun():
    print (everyLambda(2))

Solution 4: Use yield to produce lambdas lazily, avoiding the late‑binding pitfall.

def testFun():
    for i in range(4):
        yield lambda x: i*x

for everyLambda in testFun():
    print (everyLambda(2))

The final execution produces the desired sequence:

These solutions demonstrate that the interview question tests the candidate’s understanding of Python closures and the late‑binding behavior, and provides multiple ways to achieve the correct result.