Python print the value of specific parameter in the function

I only want to return and print the error when the function is called, but it seems that the whole functions are being run when I print the function.

My expected output is only:

false    
list index out of range

but I am getting:

false
false
false
list index out of range

I tried calling the function like this but did not work: print(test(error))

Question: How can I only print the error parameter and not the other parameter outside the function? Here is my code, Thanks:

def test(error=None, parameter1=None):
       array = []

       try:
              if parameter1 == True:
                     print("true")
                     array[0]
              else:
                     print("false")
                     array[0]
       except Exception as e:
              error = e
              return error
test()
if test() is not None:
       print(test())

>Solution :

You’re running the function twice, once in the if statement, and then again in the print() statement.

If you only want to run it once, assign the result to a variable.

err = test()
if not err:
    print(err)