problem is that when in sub() you give a letter instead of a number it will give a error and after that if you give both numbers it wont return anything
import math
print("Calculator")
def sub():
try:
num1 = int(input("Give a number: "))
num2 = int(input("Give a number: "))
except (TypeError, ValueError):
print("This input is invalid.")
sub()
else:
return num1,num2
def main():
while True:
num1,num2 = sub()
print("(1)+")
print("(2)-")
print("(3)*")
print("(4)/")
print("(5)sin(number1/number2)")
print("(6)cos(number1/number2)")
print("(7)Change numbers")
print("(8)Quit")
print("Current numbers: ", num1, num2)
num3 = int(input("Please select something (1-6): "))
if num3 == 1:
print("The result is:", int(num1) + int(num2))
elif num3 == 2:
print("The result is:", int(num1) - int(num2))
elif num3 == 3:
print("The result is:", int(num1) * int(num2))
elif num3 == 4:
print("The result is:", int(num1) / int(num2))
elif num3 == 5:
print("The result is:", (math.sin(int(num1) / int(num2))))
elif num3 == 6:
print("The result is:", (math.cos(int(num1) / int(num2))))
elif num3 == 7:
num1 = int(input("Give the first number: "))
num2 = int(input("Give the second number: "))
continue
elif num3 == 8:
print("Thank you!")
break
else:
print("Selection was not correct.")
if __name__ == "__main__":
main()
as in your main() function, use a while statement
def sub():
while True:
try:
num1 = int(input("Give a number: "))
num2 = int(input("Give a number: "))
except (TypeError, ValueError):
print("This input is invalid.")
# sub()
continue
else:
return num1, num2You should also account for division by zero (selections 4, 5 and 6). You do not wish to use decimal numbers?
i changed sub() to num_1() and num_2()
it works fine untill i use a letter. then i get this error
print("The result is:", int(num1) - int(num2))
TypeError: int() argument must be a string, a bytes-like object or a number, not 'NoneType'
>>>
import math
print("Calculator")
def num_1():
try:
num1 = int(input("Give a number: "))
except (TypeError, ValueError):
print("This input is invalid.")
num1 = None
num_1()
else:
return num1
def num_2():
try:
num2 = int(input("Give a number: "))
except (TypeError, ValueError):
print("This input is invalid.")
num2 = None
num_2()
else:
return num2
def main():
while True:
num1 = num_1()
num2 = num_2()
print("(1)+")
print("(2)-")
print("(3)*")
print("(4)/")
print("(5)sin(number1/number2)")
print("(6)cos(number1/number2)")
print("(7)Change numbers")
print("(8)Quit")
print("Current numbers: ", num1, num2)
num3 = int(input("Please select something (1-6): "))
if num3 == 1:
print("The result is:", int(num1) + int(num2))
elif num3 == 2:
print("The result is:", int(num1) - int(num2))
elif num3 == 3:
print("The result is:", int(num1) * int(num2))
elif num3 == 4:
print("The result is:", int(num1) / int(num2))
elif num3 == 5:
print("The result is:", (math.sin(int(num1) / int(num2))))
elif num3 == 6:
print("The result is:", (math.cos(int(num1) / int(num2))))
elif num3 == 7:
num1 = None
num2 = None
main()
elif num3 == 8:
print("Thank you!")
break
else:
print("Selection was not correct.")
if __name__ == "__main__":
main()
Your original post works fine, if you add the "while" in the sub() function. See my reply above. Also remember to comment out or better yet, delete the "sub()" within the "sub()" function as it is not needed, again, see my reply above.
yeah but i need them to be asked separately
First, you're doing the same thing twice. You have two functions that are basically identical. Make one function that gets a number, and use that function twice.
Second, follow the code. What happens when you have an invalid input? It goes into the except block, and calls itself again recursively. What does it do with the result of calling itself? Nothing. You need to return the result of the function calling itself. (return num_1())