Sunday, May 6, 2018
Function parameter types in Python
Answer
Answer
Unless I'm mistaken, creating a function in Python works like this:
def my_func(param1, param2):
# stuff
However, you don't actually give the types of those parameters. Also, if I remember, Python is a strongly typed language, as such, it seems like Python shouldn't let you pass in a parameter of a different type than the function creator expected. However, how does Python know that the user of the function is passing in the proper types? Will the program just die if it's the wrong type, assuming the function actually uses the parameter? Do you have to specify the type?
Answer
Python is strongly typed because every object has a type, every object knows its type, it's impossible to accidentally or deliberately use an object of a type "as if" it was an object of a different type, and all elementary operations on the object are delegated to its type.
This has nothing to do with names. A name in Python doesn't "have a type": if and when a name's defined, the name refers to an object, and the object does have a type (but that doesn't in fact force a type on the name: a name is a name).
A name in Python can perfectly well refer to different objects at different times (as in most programming languages, though not all) -- and there is no constraint on the name such that, if it has once referred to an object of type X, it's then forevermore constrained to refer only to other objects of type X. Constraints on names are not part of the concept of "strong typing", though some enthusiasts of static typing (where names do get constrained, and in a static, AKA compile-time, fashion, too) do misuse the term this way.
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