Python integer conversion program error question

Asked 2 years ago, Updated 2 years ago, 117 views

sel=int("Determine input number (16/10/8/2) :")

if sel != 16 or 10 or 8 or 2:
     print ("Error")

if sel == 16 or 10 or 8 or 2:
     num=input ("Enter Value:")

if sel == 16 :
     num10 = int(num,16)
if sel == 10 :
     num10 = int(num,10)
if sel == 8 :
     num10 = int(num,8)
if sel == 2 :
     num10 = int(num,2)

print ("hex==>", hex(num10))
print("decimal==>", num10)
print("Hex ==>", oct(num10))
print("Binary ==>",bin(num10))

I made a program like this and made it say 16, 8, 2, 10 or something else, but even if I type 16, it says it's an error. How do I get a program to print out an error when I put a number other than the specified number?

python base-conversion

2022-09-20 17:40

1 Answers

That's not how you use the or command. Make the following corrections.

if sel != 16 and sel != 10 and sel != 8 and sel != 2:


2022-09-20 17:40

If you have any answers or tips


© 2024 OneMinuteCode. All rights reserved.