Any way to avoid duplicate code for mut not mut reference inside struct? [duplicate]

Question

Here what I want to achieve:

trait Foo {
    fn readonly(&self) -> i32;
    fn modify(&mut self, val: i32);
}

struct FooWrapper<'a> {
    foo: &'a Foo,
}

impl<'a> FooWrapper<'a> {
    fn readonly(&self) -> i32 {
        self.foo.readonly()
    }

    fn modify(&mut self, val: i32) {
        self.foo.modify(val);//!!!
    }
}

As input I got &Foo and &mut Foo, like: fn func(a: &Foo, b: &mut Foo).

I want then wrap them inside FooWraper, and use it's methods to work with Foo.

But as you see compiler not allow code marked with //!!!.

Any way to fix this without code duplication like:

struct FooWrapper<'a> {
    foo: &'a Foo,
}
struct FooWrapperMut<'a> {
    foo: &'a mut Foo,
}
impl<'a> FooWrapper<'a>..
impl<'a> FooWrapperMut<'a>..

?

But as you see compiler not allow code — you need to include the complete compiler error. It is not acceptable to simply state that an error occurs. — Shepmaster
– Shepmaster, Commented Jun 9, 2017 at 12:43

Peter Hall · Accepted Answer · 2017-06-09 12:01:02Z

2

Just make foo a mutable reference. Then it can be borrowed either mutably or immutably.

struct FooWrapper<'a> {
    foo: &'a mut Foo,
}

A &mut T reference can always be coerced to a &T.

If you wanted to provide minimal functionality for an immutable reference and some additional methods if you able to get a mutable reference, you could maybe split it out with type parameter:

trait Foo {
    fn readonly(&self) -> i32;
    fn modify(&mut self, val: i32);
}

struct FooWrapper<T> {
    foo: T,
}

impl <'a> FooWrapper<&'a Foo> {
    fn readonly(&self) -> i32 {
        self.foo.readonly()
    }
}

impl <'a> FooWrapper<&'a mut Foo> {
    fn modify(&mut self, val: i32) {
        self.foo.modify(val); //
    }
}

Now you can supply the type parameter to be either &mut Foo or just &Foo, depending on what you have.

edited Jun 9, 2017 at 12:01

answered Jun 9, 2017 at 11:54

Peter Hall

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2 Comments

user1244932 Over a year ago

"Just make foo a mutable reference" I cann't, if I get &Foo as input how can I construct &mut Foo from it? See func as example.

Peter Hall Over a year ago

You probably need to include more code in order to explain your problem, and to show that it is different from the duplicate (if it is).

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