why is my for loop printing 2 at each iteration if b has not even been defined as a placeholder in the for loop declaration?

Question

I am fairly new to python and I just learned about tuple unpacking, so I was playing around with this concept and got a weird result.

nList = [(1,2),4,5,6]
for a in nList:
    print(a)
    print(b)

I was expecting my program to crash since b is not defined as a placeholder and even if it was only the first element of my list is a tuple, but instead, I got the following result:

you must have had a variable `b` left over in memory from something you were doing earlier. and it had a value of `2`. Do these same steps in a fresh session/kernel. — Paritosh Singh, Jul 03 '19 at 03:39
The code doesn't look correct to me. Are you sure that's your actual code? Do you have `b` define somewhere else in your script? — absolutelydevastated, Jul 03 '19 at 03:52

score 0 · Answer 1 · answered Jul 03 '19 at 03:50

You have not initialized the variable b in your code, because of which it is using some garbage value of b from memory (which is 2). To avoid such scenarios you must initialize your variables with some value (or 0) in start of your code before using them.

why is my for loop printing 2 at each iteration if b has not even been defined as a placeholder in the for loop declaration?

1 Answers1