Why is my function running into infinite loop?

I have recently began learning Python, appreciate your response on this.

I have written the definition of get_int() and get_float() differently on purpose because I wanted to understand what happens if I accept the input in main() and pass that value during function call. Here is my code:

#!/usr/bin/env python3
def get_int(prompt, low, high):
while True:
#number = float(input(prompt))
if prompt <= low or prompt > high:
print("Entry should be greater than ", low, "and less than or equal to ", high)
else:
return prompt
def get_float(prompt, low, high):
while True:
number = float(input(prompt))
if number <= low or number > high:
print("Entry should be greater than ", low, "and less than or equal to ", high)
else:
return number    
def calculate_future_value(monthly_investment, yearly_interest, years):
# convert yearly values to monthly values
monthly_interest_rate = yearly_interest / 12 / 100
months = years * 12
# calculate future value
future_value = 0.0
for i in range(months):
future_value += monthly_investment
monthly_interest = future_value * monthly_interest_rate
future_value += monthly_interest
return future_value
def main():
years = 0
monthly_investment = 0.0
yearly_interest_rate = 0.0
choice = "y"
while choice.lower() == "y":
# get input from the user
monthly_investment = get_float("Enter monthly investment: ", 0, 1000)
yearly_interest_rate = get_float("Enter yearly interest rate: ", 0, 15)
years = int(input("Enter number of years: "))
years = get_int(years, 0, 15)
# get and display future value
future_value = calculate_future_value(
monthly_investment, yearly_interest_rate, years)
print(f"Future value:\t\t\t{round(future_value, 2)}")
print()
# see if the user wants to continue
choice = input("Continue? (y/n): ")
print()
print("Bye!")
if __name__ == "__main__":
main()

I am having problem only with the get_int(). As you see, I have passed low=0 and high=15. If I input years as 0 or 16, then it displays infinite print statement specified in the "if" clause but there is no problem when I input value within the range (see screenshot: https://drive.google.com/drive/u/0/folders/1w4-IhBveZQv0pSDygc1HexNFBDUDNE-B).

I'm not entirely sure if I can use the function argument directly in the function definition without assigning to a local variable.

If it is indeed allowed, is there a mistake with my "while" loop inside get_int()? or do I include a while loop in the main() itself where I'm accepting the int value?

Thanks in advance!

while True: within the get_int is indeed an infinite loop, that while can i.e. be broken by executing a return (which you do on the else case) or a break statement. Since you don't execute any similar statement in your if, the loop continues.

As it seems by your question, you don't want a loop at all, therefore, you could just declare the get_int as:

def get_int(prompt, low, high):
#number = float(input(prompt))
if prompt <= low or prompt > high:
print("Entry should be greater than ", low, "and less than or equal to ", high)
else:
return prompt

If prompt is not correct, you don't update its value. If you don't update its value, the condition won't change. If it doesn't change, you will only print the text and never return.

Refresh prompt value if it's wrong

You still need to refresh the value somewhere if the user input is wrong. If you don't want to ask at the beginning of the loop, ask when the prompt is wrong.

def get_int(prompt, low, high):
    while True:
        if prompt <= low or prompt > high:
            print("Entry should be greater than ", low, "and less than or equal to ", high)
            prompt = int(input("Enter a new value"))  # You need to ask a new value
        else:
            return prompt

Alternative:

You can put the if condition directly in the while loop:

def get_int(prompt, low, high):
    while prompt <= low or prompt > high:
        print("Entry should be greater than ", low, "and less than or equal to ", high)
        prompt = int(input("Enter a new value"))
    return prompt

Doing so, you quit the loop right away if the prompt is correct. If it is not, you print the message and refresh the value.