> whether or not it was better to design around the fact that there could be the absence of “something” coming into existence when it should have been initialized
So this is actually why "no null, but optional types" is such a nice spot in the programming language design space. Because by default, you are making sure it "should have been initialized," that is, in Rust:
struct Point {
x: i32,
y: i32,
}
You know that x and y can never be null. You can't construct a Point without those numbers existing.
By contrast, here's a point where they could be:
struct Point {
x: Option,
y: Option,
}
You know by looking at the type if it's ever possible for something to be missing or not.
> Are there any pitfalls in Rust when Optional does not return anything?
So, Rust will require you to handle both cases. For example:
let x: Option = Some(5); // adding the type for clarity
dbg!(x + 7); // try to debug print the result
This will give you a compile-time error:
error[E0369]: cannot add `{integer}` to `Option`
--> src/main.rs:4:12
|
4 | dbg!(x + 7); // try to debug print the result
| - ^ - {integer}
| |
| Option
|
note: the foreign item type `Option` doesn't implement `Add`
It's not so much "pitfalls" exactly, but you
can choose to do the same thing you'd get in a language with null: you can choose not to handle that case:
let x: Option = Some(5); // adding the type for clarity
let result = match x {
Some(num) => num + 7,
None => panic!("we don't have a number"),
};
dbg!(result); // try to debug print the result
This will successfully print, but if we change `x` to `None`, we'll get a panic, and our current thread dies.
Because this pattern is useful, there's a method on Option called `unwrap()` that does this:
let result = x.unwrap();
And so, you can argue that Rust doesn't
truly force you to do something different here. It forces you to make an active choice, to handle it or not to handle it, and in what way. Another option, for example, is to return a default value. Here it is written out, and then with the convenience method:
let result = match x {
Some(num) => num + 7,
None => 0,
};
let result = x.unwrap_or(0);
And you have other choices, too. These are just two examples.
--------------
But to go back to the type thing for a bit, knowing statically you don't have any nulls allows you to do what some dynamic language fans call "confident coding," that is, you don't always need to be checking if something is null: you already know it isn't! This makes code more clear, and more robust.
If you take this strategy to its logical end, you arrive at "parse, don't validate," which uses Haskell examples but applies here too: https://lexi-lambda.github.io/blog/2019/11/05/parse-don-t-va...