Reputation: 49
I want to see ValueError
4
times but it is showing once, why the program cutting to search the other double numbers?
def isitDoubleorSingle(value):
if(value%2!=0):
raise ValueError("Number isn't double")
print(value)
list=[10,22,79,43,11,80]
for x in list:
isitDoubleorSingle(x)
Upvotes: 1
Views: 2641
Reputation: 389
This will solve your problem. You have to catch your Error in the except block or your script will stop running at your first raise ValueError()
edit: As @Nin17 said, you shouldn't redefine the built-in list, so renaming the list in my_list(or any name you want) should be better.
def isitDoubleorSingle(value):
try:
if(value%2!=0):
raise ValueError()
except ValueError:
print(f"Number {value} isn't double")
my_list=[10,22,79,43,11,80]
for x in my_list:
isitDoubleorSingle(x)
Upvotes: 4
Reputation: 101
When you raise an exception, the program is already closed automatically, so it is not possible to display the ValueError more than once
Upvotes: 1