The function is supposed to receive a number representing a year, and then print if it's a leap year or not.
def isItALeapYear(year):
while True:
if year % 4 == 0:
print("That is a leap year! ")
break
elif year % 4 != 0:
print("That is not a leap year...")
break
elif not isinstance(year, int) or year == None:
print("Please enter a number...")
break
The program works, the only thing I can't get right is that it is supposed to notify you if anything that it's not a number is being used as an argument. I've tried both the isinstance() function, as well as writing what I want as year != int. And then year == None in the hopes of making it work in case anything nondefined is used as an argument.
I read this post with the exact same error: TypeError: not all arguments converted during string formatting python
But I'm not intending to format anything with the % symbol As far as I'm concerned the % can be used as an operand to get the residue of a division. So in this case I'm using it to figure out if a year is a leap year or not by asking if the residue is 0 when divided by 4. I'm pretty stuck, and the sad thing is the error comes up in the very first "if", so I can't really know if the last lines for excluding any non int type argument work or not. Any help would be really appreciated!
CodePudding user response:
First thing while loop
in the function makes no sense you can remove it(If you want).
There are multiple ways to do that I will show.
First One.
def isItALeapYear(year):
if type(year) != int:
return # here return to exit the function
while True:
if year % 4 == 0:
print("That is a leap year! ")
break
elif year % 4 != 0:
print("That is not a leap year...")
break
elif not isinstance(year, int) or year == None:
print("Please enter a number...")
break
Another is.
def isItALeapYear(year):
try:
int(year)
except ValueError: # this line of code executes when the year is not the int
return # here return to exit the function
while True:
if year % 4 == 0:
print("That is a leap year! ")
break
elif year % 4 != 0:
print("That is not a leap year...")
break
elif not isinstance(year, int) or year == None:
print("Please enter a number...")
break
I know there are more ways to do that but these are the best ones (I Think).
Your function is not accurate then you can use this one.
def isItALeapYear(year):
if type(year) != int:
return
if (( year@0 == 0)or (( year%4 == 0 ) and ( year0 != 0))):
print(f"{year} is a Leap Year")
else:
print(f"{year} is Not the Leap Year")
Edit For Quetioner
def isItALeapYear(year):
if type(year) != int:
return
if (( year@0 == 0)or (( year%4 == 0 ) and ( year0 != 0))):
print(f"{year} is a Leap Year")
else:
print(f"{year} is Not the Leap Year")
try:
isItALeapYear(asdasd)
except NameError:
print("You give the wrong Value")
CodePudding user response:
I recommend to divide the functionality of checking the number from returning the output as well as from receiving the input.
def is_multiple_of_four(number: int):
if number % 4 == 0:
return True
else:
return False
if __name__ == '__main__':
user_input = ""
while not user_input.isdigit():
user_input = input("Please type in a year: ")
if is_multiple_of_four(int(user_input)):
print("That is a leap year!")
else:
print("That is not a leap year.")
Here you can see the function that checks the number does only that, it checks the number if it's modulo of 4 equals 0. In the script outside the function you can retrieve the user input for as long as it takes to get a valid numeric and return the output in respect of the functions results.
CodePudding user response:
You could use the isleap()
function from the standard library (module calendar
):
from calendar import isleap
def isItALeapYear(year):
if not isinstance(year, int):
print("Please provide a number")
elif isleap(year):
print("That is a leap year!")
else:
print("That is not a leap year...")