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Python

Posted by jamesflint00
jamesflint00
42,676 Points

TypeError: __str__() missing 1 required positional argument: 'pattern'

No sure what is going on here but when I run this in an IDE I get the following Traceback TypeError: str() missing 1 required positional argument: 'pattern'

But when I put the print in the parent class init it prints the pattern, not sure why it isn't going in the parent class init?

morse.py 
class Letter:
    def __init__(self, pattern=None):
        self.pattern = pattern

    def __str__(self, pattern):
        for index, value in enumerate(pattern):
            if value == ".":
                pattern[index] = "dot"
            elif value == "_":
                pattern[index] = "dash"
            else:
                continue
        return "-".join(pattern)

class S(Letter):
    def __init__(self):
        pattern = ['.', '_', '.']
        super().__init__(pattern)

1 Answer

Michael Hulet
Michael Hulet
47,912 Points
Michael Hulet
Michael Hulet
47,912 Points

print takes any kind of object as a parameter, but you can only print strings to the console, so that means the print function has to convert whatever it takes to a string. It does this by calling __str__ on the object it's passed, which it expects to return the object in string form. The __str__ method on all objects should generally shouldn't take any required arguments at the call site, but in your implementation of it, you're requiring that whoever calls __str__ on your Letter class (or any of its subclasses) pass it a parameter pattern. The people who wrote Python's print function don't know about your class, so they just call __str__ without parameters. In other words, print is calling __str__ on your object like this:

stringified = your_object.__str__()

But you're requiring it to call it by doing something like this:

stringified = your_object.__str__([".", "_", "."])

You're not gonna be able to talk to the developers of Python and get them to change the language to call __str__ like you want it to, but you can tell Python to not worry about that one extra parameter you're asking for. You do this by setting it to have a default value, like you did in Letter's __init__, where you set the pattern object to have a default value of None. If you do something like that on the __str__ method, print will be able to convert your object to a string without your program crashing

jamesflint00
jamesflint00
42,676 Points

Thanks Michael