"break outside loop" but i've tried everything and still got the same error message

Question

# Program make a simple calculator
# This function adds two numbers
def add(x, y):
    return x + y
# This function subtracts two numbers
def subtract(x, y):
    return x - y
# This function multiplies two numbers
def multiply(x, y):
    return x * y
# This function divides two numbers
def divide(x, y):
    return x / y
print("Select operation.")
print("1.Add")
print("2.Subtract")
print("3.Multiply")
print("4.Divide")
while True:
# take input from the user
    choice = input("Enter choice(1/2/3/4): ")
# check if choice is one of the four options
if choice in ('1', '2', '3', '4'):
    num1 = float(input("Enter first number: "))
    num2 = float(input("Enter second number: "))
if choice == '1':
    print(num1, "+", num2, "=", add(num1, num2))
elif choice == '2':
    print(num1, "-", num2, "=", subtract(num1, num2))
elif choice == '3':
    print(num1, "*", num2, "=", multiply(num1, num2))
elif choice == '4':
    print(num1, "/", num2, "=", divide(num1, num2))
# check if user wants another calculation
# break the while loop if answer is no
next_calculation = input("Let's do next calculation? (yes/no): ")
if next_calculation == "no":
    break
else:
    print("Invalid Input")

the error: File "main.py", line 38 break ^ SyntaxError: 'break' outside loop

** Process exited - Return Code: 1 ** Press Enter to exit terminal

I'm trying to make an interactive calculator, and the only issue is that I don't know how to break the loop. i tried indenting it, and it gave this. deleting the indentation made python tell me that I need to add an indent; yet adding an indent gives this error message. I don't know what else to do; I'm new to this.

Does this answer your question? [Asking the user for input until they give a valid response](https://stackoverflow.com/questions/23294658/asking-the-user-for-input-until-they-give-a-valid-response) — FlyingTeller, May 04 '22 at 16:27
https://www.programiz.com/python-programming/examples/calculator — FlyingTeller, May 04 '22 at 16:29
i decided to replace "break" with a sys.exit(), so that works I guess — slicdcheese, May 04 '22 at 16:42

score 1 · Answer 1 · answered May 04 '22 at 16:29

Try it with indentation like this:

# Program make a simple calculator
# This function adds two numbers
def add(x, y):
    return x + y
# This function subtracts two numbers
def subtract(x, y):
    return x - y
# This function multiplies two numbers
def multiply(x, y):
    return x * y
# This function divides two numbers
def divide(x, y):
    return x / y
print("Select operation.")
print("1.Add")
print("2.Subtract")
print("3.Multiply")
print("4.Divide")
while True:
    # take input from the user
    choice = input("Enter choice(1/2/3/4): ")
    # check if choice is one of the four options
    if choice in ('1', '2', '3', '4'):
        num1 = float(input("Enter first number: "))
        num2 = float(input("Enter second number: "))
    if choice == '1':
        print(num1, "+", num2, "=", add(num1, num2))
    elif choice == '2':
        print(num1, "-", num2, "=", subtract(num1, num2))
    elif choice == '3':
        print(num1, "*", num2, "=", multiply(num1, num2))
    elif choice == '4':
        print(num1, "/", num2, "=", divide(num1, num2))
    # check if user wants another calculation
    # break the while loop if answer is no
    next_calculation = input("Let's do next calculation? (yes/no): ")
    if next_calculation == "no":
        break
    else:
        print("Invalid Input")

score 0 · Answer 2 · answered May 04 '22 at 16:31

Another option would be to add a state variable like this. And move it to false when you want to exit the loop.

# Program make a simple calculator
# This function adds two numbers
def add(x, y):
    return x + y
# This function subtracts two numbers
def subtract(x, y):
    return x - y
# This function multiplies two numbers
def multiply(x, y):
    return x * y
# This function divides two numbers
def divide(x, y):
    return x / y
print("Select operation.")
print("1.Add")
print("2.Subtract")
print("3.Multiply")
print("4.Divide")
state = True
while state:
# take input from the user
    choice = input("Enter choice(1/2/3/4): ")
# check if choice is one of the four options
    if choice in ('1', '2', '3', '4'):
        num1 = float(input("Enter first number: "))
        num2 = float(input("Enter second number: "))
    if choice == '1':
        print(num1, "+", num2, "=", add(num1, num2))
    elif choice == '2':
        print(num1, "-", num2, "=", subtract(num1, num2))
    elif choice == '3':
        print(num1, "*", num2, "=", multiply(num1, num2))
    elif choice == '4':
        print(num1, "/", num2, "=", divide(num1, num2))
    # check if user wants another calculation
    # break the while loop if answer is no
    next_calculation = input("Let's do next calculation? (yes/no): ")
    if next_calculation == "no":
        state = False
    else:
        print("Invalid Input")

"break outside loop" but i've tried everything and still got the same error message

2 Answers2