Why is Python using old value of a variable? How to achieve the same using JavaScript?

Asked Nov 19 '21 at 22:52

Active Nov 19 '21 at 22:52

Viewed 75 times

I have this code which is for a LeetCode problem.

class Solution:
    def isRobotBounded(instructions: str) -> bool:
        dirX, dirY = 0, 1
        x, y = 0, 0

        for d in instructions:
            if d == "G":
                x, y = x + dirX, y + dirY
            elif d == "L":
                dirX, dirY = -1*dirY, dirX
            else:
                dirX, dirY = dirY, -1*dirX

        return (x, y) == (0, 0) or (dirX, dirY) != (0, 1)
    print(isRobotBounded('GGLLGG'))

Now, in d == "L", why is dirX used as 0 to calculate the value of dirY? In JavaScript, dirX would be calculated (-1*dirY) and then used to calculate dirY.

What am I missing here regarding Python? It looks like Python is using the initial values of the variables. What is this behavior called? And how to replicate it in JavaScript?

Thank you.

asked Nov 19 '21 at 22:52

captaincustard

1

https://stackoverflow.com/questions/6967632/unpacking-extended-unpacking-and-nested-extended-unpacking – fdireito Nov 19 '21 at 22:54
1

The right side of the assignment is evaluated completely before the variables on the left are assigned. – Barmar Nov 19 '21 at 22:56
1

Python evaluates the *whole* expression on the right before doing the assignment part of that statement. That is why `a,b=b,a` can be used to swap `a` & `b`. – Scott Hunter Nov 19 '21 at 22:57
How do I do the same in JavaScript? – captaincustard Nov 19 '21 at 22:58
2

Use destructuring: `[dirX, dirY] = [-1*dirY, dirX];` – Barmar Nov 19 '21 at 23:00
@Barmar would you like to put both your comments in an answer so I can accept it? – captaincustard Nov 19 '21 at 23:04
No, this has been answered in numerous other questions. – Barmar Nov 19 '21 at 23:04

Why is Python using old value of a variable? How to achieve the same using JavaScript?

0 Answers0