CODEWITHSUNDEEP
python-3.x
M C
M C

Reputation: 3

Problem with compound assignment operators

Im stuck with this exercise, because if i wrote with shortcut operator the result is 1, therefor is 9

    a = 6
    b = 3
    a /= 2 * b
    print(a)

a = a / 2 * b [result 9] a /= 2 * b [result 1]

BUT if i do this exercise with * and after / like this:

    a *= 2 / b

Why in this case they dont do (2/b) first?

Upvotes: 0

Views: 1748

Answers (1)

DeepSpace
DeepSpace

Reputation: 81624

a = a / 2 * b is a = 6 / 2 * 3 (following 'normal' math PEMDAS rules).

On the other hand,

a /= 2 * b is a = 6 / (2 * 3) (since the right hand side must be evaluated first, this essentially becomes a /= 6 -> a = a / 6)

why the right-hand side must be evaluated first? because the statement (a = 6 / 2) * 3 does not make sense.

Regarding your edit: The exact same behavior happens when comparing a *= 2 / b and a = a * 2 / b. The difference is that in this example a is 4 in both cases because both (6 * 2) / 3 and
6 * (2 / 3) evaluate to 4.

Upvotes: 1

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