Python Coding Challenge - Question with Answer (ID -281125)

 

Step-by-Step Explanation

✅ 1. Create an Empty List

funcs = []

This list will store functions (lambdas).

---

✅ 2. Loop Runs 3 Times

for i in range(3):

The values of i will be:

0 → 1 → 2

---

✅ 3. Lambda Is Appended Each Time

funcs.append(lambda: i)

⚠️ Important:
You are NOT storing the VALUE of i,
You are storing a function that will return i later.

So after the loop:

• funcs[0] → returns i

• funcs[1] → returns i

• funcs[2] → returns i

All three lambdas refer to the SAME variable i.

---

✅ 4. Final Value of i

After the loop finishes:

i = 2

---

✅ 5. Calling All Functions

print([f() for f in funcs])

This executes:

f() → returns i → which is 2

So all functions return:

[2, 2, 2]

---

✅ ✅ Final Output

[2, 2, 2]

---

Why This Happens? (Late Binding)

Python lambdas:

• Do NOT remember the value

• They remember the variable reference

• Value is looked up only when the function is called

This behavior is called Late Binding.

---

✅ ✅ ✅ Correct Way (Fix This Problem)

✅ Solution 1: Use Default Argument

funcs = []

for i in range(3):
    funcs.append(lambda i=i: i)
print([f() for f in funcs])

✅ Output:

[0, 1, 2]

Why this works:

• i=i captures the value immediately

• Each lambda stores its own copy

Python for Civil Engineering: Concepts, Computation & Real-world Applications