如何从Rust函数返回向量元素？

Question

I would like to return an element of a vector:

struct EntryOne {
    pub name: String,
    pub value: Option<String>,
}

struct TestVec {}

impl TestVec {
    pub fn new() -> TestVec {
        TestVec {}
    }

    pub fn findAll(&self) -> Vec<EntryOne> {
        let mut ret = Vec::new();
        ret.push(EntryOne {
            name: "foo".to_string(),
            value: Some("FooVal".to_string()),
        });
        ret.push(EntryOne {
            name: "foo2".to_string(),
            value: Some("FooVal2".to_string()),
        });
        ret.push(EntryOne {
            name: "foo3".to_string(),
            value: None,
        });
        ret.push(EntryOne {
            name: "foo4".to_string(),
            value: Some("FooVal4".to_string()),
        });

        ret
    }

    pub fn findOne(&self) -> Option<EntryOne> {
        let mut list = &self.findAll();

        if list.len() > 0 {
            println!("{} elements found", list.len());
            list.first()
        } else {
            None
        }
    }
}

fn main() {
    let test = TestVec::new();
    test.findAll();
    test.findOne();
}

( playground )

I always get this error:

error[E0308]: mismatched types
  --> src/main.rs:40:13
   |
35 |     pub fn findOne(&self) -> Option<EntryOne> {
   |                              ---------------- expected `std::option::Option<EntryOne>` because of return type
...
40 |             list.first()
   |             ^^^^^^^^^^^^ expected struct `EntryOne`, found &EntryOne
   |
   = note: expected type `std::option::Option<EntryOne>`
              found type `std::option::Option<&EntryOne>`

How do I return an element?

Answer 1

Look at the signature for Vec::first :

fn first(&self) -> Option<&T>

Given a reference to a vector, it will return a reference to the first item if there is one, and None otherwise. That means that the vector containing the values must outlive the return value, otherwise the reference would point to undefined memory.

There are two main avenues:

1. If you cannot change the vector, then you will need to make a copy of your data structure. The easiest way to do this is to annotate the structure with #[derive(Clone)] . Then you can call Option::cloned on the result of first .

2. If you can change the vector, then you can remove the first value from it and return it. There are many ways of doing this, but the shortest code-wise is to use the drain iterator.

#[derive(Debug, Clone)]
struct EntryOne {
    name: String,
    value: Option<String>,
}

fn find_all() -> Vec<EntryOne> {
    vec![
        EntryOne {
            name: "foo".to_string(),
            value: Some("FooVal".to_string()),
        },
        EntryOne {
            name: "foo2".to_string(),
            value: Some("FooVal2".to_string()),
        },
        EntryOne {
            name: "foo3".to_string(),
            value: None,
        },
        EntryOne {
            name: "foo4".to_string(),
            value: Some("FooVal4".to_string()),
        },
    ]
}

fn find_one_by_clone() -> Option<EntryOne> {
    find_all().first().cloned()
}

fn find_one_by_drain() -> Option<EntryOne> {
    let mut all = find_all();
    let mut i = all.drain(0..1);
    i.next()
}

fn main() {
    println!("{:?}", find_one_by_clone());
    println!("{:?}", find_one_by_drain());
}

Additional changes:

1. There's no need for TestVec if there's no state; just make functions.
2. Rust style is snake_case for method and variable names.
3. Use vec! to construct a vector when providing all the elements.
4. Derive Debug so you can print the value.

---

If you wanted to always get the last element, you can use pop :

fn find_one_by_pop() -> Option<EntryOne> {
    find_all().pop()
}