Don't understand why it works

pub fn do_find(a_sequence:&str,a_str: &str)->String
match a_sequence.split_whitespace().find(
|&word|{word == a_str}//Here, passing &word makes sense to me but passing just word doesn't make sense as find takes its operand by ref, yet I'm allowed to pass just word and it still compiles

Basically what I don't understand is this:
If I have a function:

fn do_something(a_str: &String)

I have to call this function like so:


But with find in the example, even though find's closure takes its arg by ref I can pass that argument by value and it still works. In my opinion this is somewhat confusing.

When you write |&word|, that is a pattern, not a type annotation. It would be like writing this:

fn do_something(&a_str: &String) {}

This is valid syntax, but probably doesn't mean what you think it does. When an ampersand is found in a pattern, that actually means to dereference the pointer.

It's just like if you want the n from the pattern &n given the value &5, then that results in n = 5 with n not being a reference.

The reason this makes sense with find is that find only provides a borrow to the closure, and since the item type is already &str, you will get &&str from find, but your pattern dereferences once, so word has type &str.

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Hi and thank you for the reply. What &a_str means in the example above? Is it a dereferencing passed pointer? Could you provide example of usage? It will be easier to understand.

Ok, I'll try to explain another time. The &a_str part is a pattern, so the type of &a_str is &String, and hence the type of a_str is String. So it means to dereference the provided reference. In this case it wont compile because String is not Copy, but with an integer it works:

fn do_something(&an_int: &u32) {
    // can be passed directly as u32
    // to pass as &u32, we need an ampersand

fn takes_int(int: u32) {
    println!("{}", int);
fn takes_ref_int(int: &u32) {
    println!("{}", int);

Patterns and Expressions are not the same; they are "opposite" constructs.

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