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0

My code looks like this:

pub enum Cache<'a, T> 
 Pending(&'a dyn FnOnce() -> T),
 Cached(T),


impl<'a, T> Cache<'a, T> 
 pub fn get(&self) -> &mut T 
 // This caches and borrows the T
 


impl<'a, T> PartialEq for Cache<'a, T>
 where &'a mut T: PartialEq 

 fn eq(&self, other: &Self) -> bool 
 self.get().eq(other.get())
 

But implementing Eq fails with:

error[E0308]: mismatched types
--> src/lib.rs:20:23
|
20 | self.get().eq(other.get())
| ^^^^^^^^^^^ expected mutable reference, found type parameter
|
= note: expected type `&&'a mut T`
 found type `&mut T`

I think I'm conceptually misunderstanding something.

types rust borrow-checker borrowing

share|improve this question

edited Mar 29 at 7:11

hellow

7,11655 gold badges2727 silver badges4848 bronze badges

asked Mar 28 at 21:10

MarkMark

10.2k55 gold badges6969 silver badges9898 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Are you trying to compare the T objects themselves, or the references returned by .get()?
  
  – Frxstrem
  Mar 28 at 21:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Frxstrem Functionally, I want the T objects to be the same.
  
  – Mark
  Mar 28 at 21:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  &'a dyn FnOnce() -> T looks like a questionable type, since you can't ever call the function. (You need to own an FnOnce to be able to call it.)
  
  – Sven Marnach
  Mar 28 at 21:51
  

  
  
  
  

add a comment
 | 

0

My code looks like this:

pub enum Cache<'a, T> 
 Pending(&'a dyn FnOnce() -> T),
 Cached(T),


impl<'a, T> Cache<'a, T> 
 pub fn get(&self) -> &mut T 
 // This caches and borrows the T
 


impl<'a, T> PartialEq for Cache<'a, T>
 where &'a mut T: PartialEq 

 fn eq(&self, other: &Self) -> bool 
 self.get().eq(other.get())
 

But implementing Eq fails with:

error[E0308]: mismatched types
--> src/lib.rs:20:23
|
20 | self.get().eq(other.get())
| ^^^^^^^^^^^ expected mutable reference, found type parameter
|
= note: expected type `&&'a mut T`
 found type `&mut T`

I think I'm conceptually misunderstanding something.

types rust borrow-checker borrowing

share|improve this question

edited Mar 29 at 7:11

hellow

7,11655 gold badges2727 silver badges4848 bronze badges

asked Mar 28 at 21:10

MarkMark

10.2k55 gold badges6969 silver badges9898 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Are you trying to compare the T objects themselves, or the references returned by .get()?
  
  – Frxstrem
  Mar 28 at 21:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Frxstrem Functionally, I want the T objects to be the same.
  
  – Mark
  Mar 28 at 21:29
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  &'a dyn FnOnce() -> T looks like a questionable type, since you can't ever call the function. (You need to own an FnOnce to be able to call it.)
  
  – Sven Marnach
  Mar 28 at 21:51
  

  
  
  
  

add a comment
 | 

My code looks like this:

pub enum Cache<'a, T> 
 Pending(&'a dyn FnOnce() -> T),
 Cached(T),


impl<'a, T> Cache<'a, T> 
 pub fn get(&self) -> &mut T 
 // This caches and borrows the T
 


impl<'a, T> PartialEq for Cache<'a, T>
 where &'a mut T: PartialEq 

 fn eq(&self, other: &Self) -> bool 
 self.get().eq(other.get())
 

But implementing Eq fails with:

error[E0308]: mismatched types
--> src/lib.rs:20:23
|
20 | self.get().eq(other.get())
| ^^^^^^^^^^^ expected mutable reference, found type parameter
|
= note: expected type `&&'a mut T`
 found type `&mut T`

I think I'm conceptually misunderstanding something.

types rust borrow-checker borrowing

share|improve this question

edited Mar 29 at 7:11

hellow

7,11655 gold badges2727 silver badges4848 bronze badges

asked Mar 28 at 21:10

MarkMark

10.2k55 gold badges6969 silver badges9898 bronze badges

pub enum Cache<'a, T> 
 Pending(&'a dyn FnOnce() -> T),
 Cached(T),


impl<'a, T> Cache<'a, T> 
 pub fn get(&self) -> &mut T 
 // This caches and borrows the T
 


impl<'a, T> PartialEq for Cache<'a, T>
 where &'a mut T: PartialEq 

 fn eq(&self, other: &Self) -> bool 
 self.get().eq(other.get())
 

error[E0308]: mismatched types
--> src/lib.rs:20:23
|
20 | self.get().eq(other.get())
| ^^^^^^^^^^^ expected mutable reference, found type parameter
|
= note: expected type `&&'a mut T`
 found type `&mut T`

share|improve this question

edited Mar 29 at 7:11

hellow

7,11655 gold badges2727 silver badges4848 bronze badges

asked Mar 28 at 21:10

MarkMark

10.2k55 gold badges6969 silver badges9898 bronze badges

share|improve this question

edited Mar 29 at 7:11

hellow

7,11655 gold badges2727 silver badges4848 bronze badges

asked Mar 28 at 21:10

MarkMark

10.2k55 gold badges6969 silver badges9898 bronze badges

edited Mar 29 at 7:11

hellow

7,11655 gold badges2727 silver badges4848 bronze badges

edited Mar 29 at 7:11

hellow

7,11655 gold badges2727 silver badges4848 bronze badges

asked Mar 28 at 21:10

MarkMark

10.2k55 gold badges6969 silver badges9898 bronze badges

asked Mar 28 at 21:10

MarkMark

10.2k55 gold badges6969 silver badges9898 bronze badges

1 Answer
1

active

oldest

votes

4

You can understand why Rust is expecting an &&mut T by looking at the definition of the eq() method in the PartialEq trait:

fn eq(&self, other: &Rhs) -> bool;

The types of the parameters to this method are &Self and &Rhs; since Rhs defaults to Self and you did not specify anything else in your trait bound, both arguments are expected to be of type &Self.

Now what is Self in this case? Your trait bound is this:

&'a mut T: PartialEq

So the only PartialEq implementation the compiler can use is the one for the type &'a mut T, so this is what Self is; &Self, in turn, must be &&'a mut T, which is exactly what the compiler is expecting.

You probably want the trait bound for T instead:

impl<'a, T> PartialEq for Cache<'a, T>
where
 T: PartialEq,

 fn eq(&self, other: &Self) -> bool 
 self.get() == other.get()
 

Also note that you can simply use == instead of explicitly calling eq(). It makes getting the types right a bit easier, since the compiler will take references of the arguments implicitly – a == b expands to PartialEq::eq(&a, &b).

share|improve this answer

answered Mar 28 at 22:09

Sven MarnachSven Marnach

385k8585 gold badges777777 silver badges716716 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I had added &'a mut based on a compiler hint I got when I tried self.get().eq(other.get()). But now I see I should have used &other instead (or easier, ==).
  
  – Mark
  Mar 29 at 20:54
  

  
  
  
  

add a comment
 | 

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4

You can understand why Rust is expecting an &&mut T by looking at the definition of the eq() method in the PartialEq trait:

fn eq(&self, other: &Rhs) -> bool;

The types of the parameters to this method are &Self and &Rhs; since Rhs defaults to Self and you did not specify anything else in your trait bound, both arguments are expected to be of type &Self.

Now what is Self in this case? Your trait bound is this:

&'a mut T: PartialEq

So the only PartialEq implementation the compiler can use is the one for the type &'a mut T, so this is what Self is; &Self, in turn, must be &&'a mut T, which is exactly what the compiler is expecting.

You probably want the trait bound for T instead:

impl<'a, T> PartialEq for Cache<'a, T>
where
 T: PartialEq,

 fn eq(&self, other: &Self) -> bool 
 self.get() == other.get()
 

Also note that you can simply use == instead of explicitly calling eq(). It makes getting the types right a bit easier, since the compiler will take references of the arguments implicitly – a == b expands to PartialEq::eq(&a, &b).

share|improve this answer

answered Mar 28 at 22:09

Sven MarnachSven Marnach

385k8585 gold badges777777 silver badges716716 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I had added &'a mut based on a compiler hint I got when I tried self.get().eq(other.get()). But now I see I should have used &other instead (or easier, ==).
  
  – Mark
  Mar 29 at 20:54
  

  
  
  
  

add a comment
 | 

4

You can understand why Rust is expecting an &&mut T by looking at the definition of the eq() method in the PartialEq trait:

fn eq(&self, other: &Rhs) -> bool;

The types of the parameters to this method are &Self and &Rhs; since Rhs defaults to Self and you did not specify anything else in your trait bound, both arguments are expected to be of type &Self.

Now what is Self in this case? Your trait bound is this:

&'a mut T: PartialEq

So the only PartialEq implementation the compiler can use is the one for the type &'a mut T, so this is what Self is; &Self, in turn, must be &&'a mut T, which is exactly what the compiler is expecting.

You probably want the trait bound for T instead:

impl<'a, T> PartialEq for Cache<'a, T>
where
 T: PartialEq,

 fn eq(&self, other: &Self) -> bool 
 self.get() == other.get()
 

Also note that you can simply use == instead of explicitly calling eq(). It makes getting the types right a bit easier, since the compiler will take references of the arguments implicitly – a == b expands to PartialEq::eq(&a, &b).

share|improve this answer

answered Mar 28 at 22:09

Sven MarnachSven Marnach

385k8585 gold badges777777 silver badges716716 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I had added &'a mut based on a compiler hint I got when I tried self.get().eq(other.get()). But now I see I should have used &other instead (or easier, ==).
  
  – Mark
  Mar 29 at 20:54
  

  
  
  
  

add a comment
 | 

You can understand why Rust is expecting an &&mut T by looking at the definition of the eq() method in the PartialEq trait:

fn eq(&self, other: &Rhs) -> bool;

The types of the parameters to this method are &Self and &Rhs; since Rhs defaults to Self and you did not specify anything else in your trait bound, both arguments are expected to be of type &Self.

Now what is Self in this case? Your trait bound is this:

&'a mut T: PartialEq

So the only PartialEq implementation the compiler can use is the one for the type &'a mut T, so this is what Self is; &Self, in turn, must be &&'a mut T, which is exactly what the compiler is expecting.

You probably want the trait bound for T instead:

impl<'a, T> PartialEq for Cache<'a, T>
where
 T: PartialEq,

 fn eq(&self, other: &Self) -> bool 
 self.get() == other.get()
 

Also note that you can simply use == instead of explicitly calling eq(). It makes getting the types right a bit easier, since the compiler will take references of the arguments implicitly – a == b expands to PartialEq::eq(&a, &b).

share|improve this answer

answered Mar 28 at 22:09

Sven MarnachSven Marnach

385k8585 gold badges777777 silver badges716716 bronze badges

fn eq(&self, other: &Rhs) -> bool;

&'a mut T: PartialEq

impl<'a, T> PartialEq for Cache<'a, T>
where
 T: PartialEq,

 fn eq(&self, other: &Self) -> bool 
 self.get() == other.get()
 

share|improve this answer

answered Mar 28 at 22:09

Sven MarnachSven Marnach

385k8585 gold badges777777 silver badges716716 bronze badges

answered Mar 28 at 22:09

Sven MarnachSven Marnach

385k8585 gold badges777777 silver badges716716 bronze badges

answered Mar 28 at 22:09

Sven MarnachSven Marnach

385k8585 gold badges777777 silver badges716716 bronze badges