Python while loop not stopping when element is found in a tuple [closed]

You never tell the loop to stop. After printing the message you still increment i and the while keeps running until the end of the tuple.

Minimal fix for your code:

numbers = (2, 4, 6, 8, 10, 12)

x = int(input("Enter a number to search: "))

i = 0
while i < len(numbers):
    if numbers[i] == x:
        print("Number found at index", i)
        break          # stop the loop as soon as we find the element
    i += 1
else:
    # this runs only if the loop finished without hitting `break`
    print("Number not found")

More Pythonic approach – use a for loop (or index) instead of managing i yourself:

numbers = (2, 4, 6, 8, 10, 12)

x = int(input("Enter a number to search: "))

for i, value in enumerate(numbers):
    if value == x:
        print("Number found at index", i)
        break
else:
    print("Number not found")

Or just:

numbers = (2, 4, 6, 8, 10, 12)
x = int(input("Enter a number to search: "))

try:
    i = numbers.index(x)
    print("Number found at index", i)
except ValueError:
    print("Number not found")

Use break to exit the loop as soon as the element is found

numbers = (2, 4, 6, 8, 10, 12)

x = int(input("Enter a number to search: "))

i = 0

while i < len(numbers):
    if numbers[i] == x:
        print("Number found at index", i)
        break   # stop the loop
    i += 1

You have to specify in the loop that the element is found you can stop now, to solve the issue use break before incrementing the value of i

Your loop condition never becomes False -- This will print “Found” but keep looping, because you never told it to stop.

t = (3, 7, 10, 1)
i = 0

while i < len(t):
    if t[i] == 10:
        print("Found")
    i += 1