Unreachable code in if statement inside a while loop - python-3.x

Question

I was trying to make a menu with a while loop to keep repeating until a valid input was entered. This is the code that I have:
print("1) Play game")
print("")
print("2) Instructions")
print("")
print("3) Quit")
print("")
while True:
try:
menu = int(input("Make a selection: "))
except ValueError:
print("Sorry, your input was not valid")
continue
if menu > 3 or menu < 1:
print("Please use a number in the range 1-3")
continue
else:
break
if menu == 1:
print("Lets go!")
elif menu == 2:
print("Lets go!")
else:
print("Quitting...")
The "if" where it says "if menu > 3 or menu < 1:" gives an error saying that:
"Unreachable codepylint(unreachable)"
Please help, it would be much appreciated.

Answer 1

Problem 1: continue statement
To help you understand your code, i used a flowchart to draw the lines for you.
However, if you move the continue inside the except statement, then the code will pass onto the if statement on a successful try statement.
The challenge is, that the code is never going into the if menu > 3.... statement, it always goes to try or except and then does the continue. So its looping itself skipping the if menu > 3 .... statement. That means, you never get out of the loop as it never gets to execute break.
Problem 2: You said you removed the continue statement.
When you remove continue from the statement, you create a new problem. If the code goes into except clause, variable menu did not get created properly. So when it goes to if menu > 3 ..., the variable has not been defined (due to except clause), and that gives you a new error.
Solution:
I would recommend to move the continue statement inside the except clause. That will take care of the problem. It will go back to while statement if there is an error with the input for menu. If it is successful, then your if menu > 3 ... will execute properly.
The modified correct code will be:
while True:
try:
menu = int(input("Make a selection: "))
except ValueError:
print("Sorry, your input was not valid")
continue #move this inside except
if menu > 3 or menu < 1:
print("Please use a number in the range 1-3")
continue
else:
break
if menu == 1:
print("Lets go!")
elif menu == 2:
print("Lets go!")
else:
print("Quitting...")