为什么即使Pattern同时具有＆str和char的实现，Split类型也只返回＆str？

Question

I'm having a hard time understanding how the Split type works in Rust.

Split<'a, P> where P: Pattern<'a> is the type returned by the std::string::String::split method. The type has an implementation for Iterator<'a, P> , where P is still the Pattern type, but in reality (and as I would expect) the Iterator only returns &str slices.

Eg, split(p).collect::<Vec<&str>>() works but split(p).collect::<Vec<char>>() results a compiler error. This is what I would expect to happen, but I don't understand how it happens since Pattern has implementations for both &str and char .

Why isn't the Split type simply defined as Split<'a, &'a str> , since it is effectively an Iterator over &str s? Why does it behave as if it is effectively defined as that?

Answer 1

The type has an implementation for Iterator<'a, P>

It doesn't. It's just Iterator , which has no type parameters. Every implementation of Iterator must declare the type of the item that it iterates over using an associated type. For example, Split 's implementation ^[1] is something like this:

impl <'a, P> Iterator for Split<'a, P> {
    type Item = &'a str;
    fn next(&mut self) -> Option<&'a str> { ... }
}

Why isn't the Split type simply defined as Split<'a, &'a str> , since it is effectively an Iterator over &str s?

Because iterators are lazy. The Split struct still needs to know about the pattern in order to match the next item. Its iterator instance has Item = &str , because that is what it iterates over.

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^[1] The actual implementation is generated by a macro .