why does my code multiply 2 times instead of one

Question

My code should verify if a number is even, if it is, it prints the number multiplied by 2, if it isn’t, it should print the number multiplied by 3, but just doesn’t work.

m = int(input())
for i in range(m):
    n = int(input())
    n*=2 if n%2==0 else n*3
    print(n)

When i try this input:

It returns:

3
4
**27** <- ?

Asked By: noobprogrammer

||

Source

Answer 1

n *= 2 if n % 2 == 0 else n * 3

means

n *= (2 if n % 2 == 0 else n * 3)

which means

if n % 2 == 0:
    n = n * 2
else:
    n = n * n * 3

You meant to write

n *= 2 if n % 2 == 0 else 3

Answered By: DeepSpace

Answer 2

n*=2 if n%2==0 else n*3

Operator precedence.

This statement is interpreted as

n *= (2 if n%2==0 else n*3)

And for input 3, n%2==0 is not true, so the statement becomes

n *= 9

Which is 27.

Answered By: John Gordon

why does my code multiply 2 times instead of one

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