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Make function parameters optional

Time:05-10

I want the function to return a value when there is an error. I created this sample code, but it has two parameters. How can I make the function parameters optional? Because when I run the code below, my IDE returns an error that the parameters are required.

def test(error, parameter1):
       if parameter1 == True:
              print("true")
       else:
              print("false")
       
       array = []
       try:
              array[0]
       except Exception as e:
              error = e
              return error

value1 = 1
value2 = 2

if value1 == 1:
       #I want this to print the error of the function above
       print(test(error))

if value2 == 2:
       # I want this to pass parameter to the function above
       test(parameter1=True)

CodePudding user response:

Optional paramteters are declared with =None, i.e.:

def test(error=None, parameter1=None):
    ...

in your function you'll have to check if the parameters are not none, i.e.:

    if error is not None:
        ....

CodePudding user response:

You could set the function parameters some default values so when no parameters are passed, the default values become the values of the variables:

def test(error = None, parameter1 = None):
       if parameter1 == True:
              print("true")
       else:
              print("false")
       
       array = []
       try:
              array[0]
       except Exception as e:
              error = e
              return error

Here None becomes a default value and if a parameter is passed, the value changes from None to the given value.

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