Why is calling a FnOnce closure a move?










7














I'm trying to pass in a closure to a function that will then mutate something passed into it within the scope of the function. Based on my current understanding of Rust, that should look something like this:



pub fn call_something(callback: &FnOnce(&mut Vec<i32>)) 
let mut my_vec = vec![0, 1, 2, 3, 4];
callback(&mut my_vec);



That results in these errors:



error[E0161]: cannot move a value of type dyn for<'r> std::ops::FnOnce(&'r mut std::vec::Vec<i32>): the size of dyn for<'r> std::ops::FnOnce(&'r mut std::vec::Vec<i32>) cannot be statically determined
--> src/lib.rs:3:5
|
3 | callback(&mut my_vec);
| ^^^^^^^^

error[E0507]: cannot move out of borrowed content
--> src/lib.rs:3:5
|
3 | callback(&mut my_vec);
| ^^^^^^^^ cannot move out of borrowed content


Why is calling a FnOnce a move? What am I missing here?










share|improve this question




























    7














    I'm trying to pass in a closure to a function that will then mutate something passed into it within the scope of the function. Based on my current understanding of Rust, that should look something like this:



    pub fn call_something(callback: &FnOnce(&mut Vec<i32>)) 
    let mut my_vec = vec![0, 1, 2, 3, 4];
    callback(&mut my_vec);



    That results in these errors:



    error[E0161]: cannot move a value of type dyn for<'r> std::ops::FnOnce(&'r mut std::vec::Vec<i32>): the size of dyn for<'r> std::ops::FnOnce(&'r mut std::vec::Vec<i32>) cannot be statically determined
    --> src/lib.rs:3:5
    |
    3 | callback(&mut my_vec);
    | ^^^^^^^^

    error[E0507]: cannot move out of borrowed content
    --> src/lib.rs:3:5
    |
    3 | callback(&mut my_vec);
    | ^^^^^^^^ cannot move out of borrowed content


    Why is calling a FnOnce a move? What am I missing here?










    share|improve this question


























      7












      7








      7







      I'm trying to pass in a closure to a function that will then mutate something passed into it within the scope of the function. Based on my current understanding of Rust, that should look something like this:



      pub fn call_something(callback: &FnOnce(&mut Vec<i32>)) 
      let mut my_vec = vec![0, 1, 2, 3, 4];
      callback(&mut my_vec);



      That results in these errors:



      error[E0161]: cannot move a value of type dyn for<'r> std::ops::FnOnce(&'r mut std::vec::Vec<i32>): the size of dyn for<'r> std::ops::FnOnce(&'r mut std::vec::Vec<i32>) cannot be statically determined
      --> src/lib.rs:3:5
      |
      3 | callback(&mut my_vec);
      | ^^^^^^^^

      error[E0507]: cannot move out of borrowed content
      --> src/lib.rs:3:5
      |
      3 | callback(&mut my_vec);
      | ^^^^^^^^ cannot move out of borrowed content


      Why is calling a FnOnce a move? What am I missing here?










      share|improve this question















      I'm trying to pass in a closure to a function that will then mutate something passed into it within the scope of the function. Based on my current understanding of Rust, that should look something like this:



      pub fn call_something(callback: &FnOnce(&mut Vec<i32>)) 
      let mut my_vec = vec![0, 1, 2, 3, 4];
      callback(&mut my_vec);



      That results in these errors:



      error[E0161]: cannot move a value of type dyn for<'r> std::ops::FnOnce(&'r mut std::vec::Vec<i32>): the size of dyn for<'r> std::ops::FnOnce(&'r mut std::vec::Vec<i32>) cannot be statically determined
      --> src/lib.rs:3:5
      |
      3 | callback(&mut my_vec);
      | ^^^^^^^^

      error[E0507]: cannot move out of borrowed content
      --> src/lib.rs:3:5
      |
      3 | callback(&mut my_vec);
      | ^^^^^^^^ cannot move out of borrowed content


      Why is calling a FnOnce a move? What am I missing here?







      rust closures move-semantics






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 9 at 22:08









      Shepmaster

      147k11281415




      147k11281415










      asked Nov 9 at 22:01









      user96425

      382




      382






















          1 Answer
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          7















          Why is calling a FnOnce a move?




          Because that's the definition of what makes a closure FnOnce:



          extern "rust-call" fn call_once(self, args: Args) -> Self::Output
          // ^^^^


          Contrast this to FnMut and Fn:



          extern "rust-call" fn call_mut(&mut self, args: Args) -> Self::Output
          // ^^^^^^^^^




          extern "rust-call" fn call(&self, args: Args) -> Self::Output
          // ^^^^^


          See also:



          • When does a closure implement Fn, FnMut and FnOnce?

          • "cannot move a value of type FnOnce" when moving a boxed function

          • Cannot move out of borrowed content


          You probably want



          pub fn call_something(callback: impl FnOnce(&mut Vec<i32>))


          or



          pub fn call_something<F>(callback: F)
          where
          F: FnOnce(&mut Vec<i32>),


          These are identical. They both take ownership of the closure, which means that you can call the closure and consume it in the process.






          share|improve this answer






















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            active

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            7















            Why is calling a FnOnce a move?




            Because that's the definition of what makes a closure FnOnce:



            extern "rust-call" fn call_once(self, args: Args) -> Self::Output
            // ^^^^


            Contrast this to FnMut and Fn:



            extern "rust-call" fn call_mut(&mut self, args: Args) -> Self::Output
            // ^^^^^^^^^




            extern "rust-call" fn call(&self, args: Args) -> Self::Output
            // ^^^^^


            See also:



            • When does a closure implement Fn, FnMut and FnOnce?

            • "cannot move a value of type FnOnce" when moving a boxed function

            • Cannot move out of borrowed content


            You probably want



            pub fn call_something(callback: impl FnOnce(&mut Vec<i32>))


            or



            pub fn call_something<F>(callback: F)
            where
            F: FnOnce(&mut Vec<i32>),


            These are identical. They both take ownership of the closure, which means that you can call the closure and consume it in the process.






            share|improve this answer



























              7















              Why is calling a FnOnce a move?




              Because that's the definition of what makes a closure FnOnce:



              extern "rust-call" fn call_once(self, args: Args) -> Self::Output
              // ^^^^


              Contrast this to FnMut and Fn:



              extern "rust-call" fn call_mut(&mut self, args: Args) -> Self::Output
              // ^^^^^^^^^




              extern "rust-call" fn call(&self, args: Args) -> Self::Output
              // ^^^^^


              See also:



              • When does a closure implement Fn, FnMut and FnOnce?

              • "cannot move a value of type FnOnce" when moving a boxed function

              • Cannot move out of borrowed content


              You probably want



              pub fn call_something(callback: impl FnOnce(&mut Vec<i32>))


              or



              pub fn call_something<F>(callback: F)
              where
              F: FnOnce(&mut Vec<i32>),


              These are identical. They both take ownership of the closure, which means that you can call the closure and consume it in the process.






              share|improve this answer

























                7












                7








                7







                Why is calling a FnOnce a move?




                Because that's the definition of what makes a closure FnOnce:



                extern "rust-call" fn call_once(self, args: Args) -> Self::Output
                // ^^^^


                Contrast this to FnMut and Fn:



                extern "rust-call" fn call_mut(&mut self, args: Args) -> Self::Output
                // ^^^^^^^^^




                extern "rust-call" fn call(&self, args: Args) -> Self::Output
                // ^^^^^


                See also:



                • When does a closure implement Fn, FnMut and FnOnce?

                • "cannot move a value of type FnOnce" when moving a boxed function

                • Cannot move out of borrowed content


                You probably want



                pub fn call_something(callback: impl FnOnce(&mut Vec<i32>))


                or



                pub fn call_something<F>(callback: F)
                where
                F: FnOnce(&mut Vec<i32>),


                These are identical. They both take ownership of the closure, which means that you can call the closure and consume it in the process.






                share|improve this answer















                Why is calling a FnOnce a move?




                Because that's the definition of what makes a closure FnOnce:



                extern "rust-call" fn call_once(self, args: Args) -> Self::Output
                // ^^^^


                Contrast this to FnMut and Fn:



                extern "rust-call" fn call_mut(&mut self, args: Args) -> Self::Output
                // ^^^^^^^^^




                extern "rust-call" fn call(&self, args: Args) -> Self::Output
                // ^^^^^


                See also:



                • When does a closure implement Fn, FnMut and FnOnce?

                • "cannot move a value of type FnOnce" when moving a boxed function

                • Cannot move out of borrowed content


                You probably want



                pub fn call_something(callback: impl FnOnce(&mut Vec<i32>))


                or



                pub fn call_something<F>(callback: F)
                where
                F: FnOnce(&mut Vec<i32>),


                These are identical. They both take ownership of the closure, which means that you can call the closure and consume it in the process.







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited Nov 9 at 22:39

























                answered Nov 9 at 22:09









                Shepmaster

                147k11281415




                147k11281415



























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