How can I move a value out of the argument to Drop::drop()?









up vote
7
down vote

favorite












I'm using gfx-hal, which requires me to create resources which need to be explicitly destroyed using functions specific to their type. I'd like to store instances of these types in structs, and I'd also like to tie cleaning them up to the lifetime of the owning struct, instead of managing their lifetimes manually and potentially having objects on the GPU/in the driver live forever.



However, all the functions in the destroy family of functions take the type directly, rather than a reference, so when I try to pass them from my structs, I get errors like the following:



error[E0509]: cannot move out of type `S`, which implements the `Drop` trait
--> src/lib.rs:9:18
|
9 | destroyT(self.member)
| ^^^^^^^^^^^ cannot move out of here


It seems like there should be some way around this issue, as I'm currently in the Drop::drop function itself, so self is already "consumed." How do I get the instances of these types out of self as T, and not &T?



struct T;

struct S
member: T,


impl Drop for S
fn drop(&mut self)
destroyT(self.member)



// elsewhere, in a library

fn destroyT(t: T)
//...










share|improve this question



















  • 2




    Looks like you're not the only one to find this frustrating: github.com/gfx-rs/gfx/issues/2452
    – loganfsmyth
    Nov 12 at 2:07










  • Couldn't you use a NewType for T that implements Drop and calls destroy(). That way, the Drop for S would be automatically generated.
    – rodrigo
    Nov 12 at 10:29






  • 1




    @rodrigo Isn't that exactly what the OP was trying to do?
    – Sven Marnach
    Nov 12 at 10:52










  • @SvenMarnach: Oh, I see. I was assuming S was some composite type. But then the NewType would be exactly like this S.
    – rodrigo
    Nov 12 at 11:02














up vote
7
down vote

favorite












I'm using gfx-hal, which requires me to create resources which need to be explicitly destroyed using functions specific to their type. I'd like to store instances of these types in structs, and I'd also like to tie cleaning them up to the lifetime of the owning struct, instead of managing their lifetimes manually and potentially having objects on the GPU/in the driver live forever.



However, all the functions in the destroy family of functions take the type directly, rather than a reference, so when I try to pass them from my structs, I get errors like the following:



error[E0509]: cannot move out of type `S`, which implements the `Drop` trait
--> src/lib.rs:9:18
|
9 | destroyT(self.member)
| ^^^^^^^^^^^ cannot move out of here


It seems like there should be some way around this issue, as I'm currently in the Drop::drop function itself, so self is already "consumed." How do I get the instances of these types out of self as T, and not &T?



struct T;

struct S
member: T,


impl Drop for S
fn drop(&mut self)
destroyT(self.member)



// elsewhere, in a library

fn destroyT(t: T)
//...










share|improve this question



















  • 2




    Looks like you're not the only one to find this frustrating: github.com/gfx-rs/gfx/issues/2452
    – loganfsmyth
    Nov 12 at 2:07










  • Couldn't you use a NewType for T that implements Drop and calls destroy(). That way, the Drop for S would be automatically generated.
    – rodrigo
    Nov 12 at 10:29






  • 1




    @rodrigo Isn't that exactly what the OP was trying to do?
    – Sven Marnach
    Nov 12 at 10:52










  • @SvenMarnach: Oh, I see. I was assuming S was some composite type. But then the NewType would be exactly like this S.
    – rodrigo
    Nov 12 at 11:02












up vote
7
down vote

favorite









up vote
7
down vote

favorite











I'm using gfx-hal, which requires me to create resources which need to be explicitly destroyed using functions specific to their type. I'd like to store instances of these types in structs, and I'd also like to tie cleaning them up to the lifetime of the owning struct, instead of managing their lifetimes manually and potentially having objects on the GPU/in the driver live forever.



However, all the functions in the destroy family of functions take the type directly, rather than a reference, so when I try to pass them from my structs, I get errors like the following:



error[E0509]: cannot move out of type `S`, which implements the `Drop` trait
--> src/lib.rs:9:18
|
9 | destroyT(self.member)
| ^^^^^^^^^^^ cannot move out of here


It seems like there should be some way around this issue, as I'm currently in the Drop::drop function itself, so self is already "consumed." How do I get the instances of these types out of self as T, and not &T?



struct T;

struct S
member: T,


impl Drop for S
fn drop(&mut self)
destroyT(self.member)



// elsewhere, in a library

fn destroyT(t: T)
//...










share|improve this question















I'm using gfx-hal, which requires me to create resources which need to be explicitly destroyed using functions specific to their type. I'd like to store instances of these types in structs, and I'd also like to tie cleaning them up to the lifetime of the owning struct, instead of managing their lifetimes manually and potentially having objects on the GPU/in the driver live forever.



However, all the functions in the destroy family of functions take the type directly, rather than a reference, so when I try to pass them from my structs, I get errors like the following:



error[E0509]: cannot move out of type `S`, which implements the `Drop` trait
--> src/lib.rs:9:18
|
9 | destroyT(self.member)
| ^^^^^^^^^^^ cannot move out of here


It seems like there should be some way around this issue, as I'm currently in the Drop::drop function itself, so self is already "consumed." How do I get the instances of these types out of self as T, and not &T?



struct T;

struct S
member: T,


impl Drop for S
fn drop(&mut self)
destroyT(self.member)



// elsewhere, in a library

fn destroyT(t: T)
//...







rust






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 12 at 1:40









Shepmaster

146k11281413




146k11281413










asked Nov 12 at 0:33









Ben Pious

3,83111526




3,83111526







  • 2




    Looks like you're not the only one to find this frustrating: github.com/gfx-rs/gfx/issues/2452
    – loganfsmyth
    Nov 12 at 2:07










  • Couldn't you use a NewType for T that implements Drop and calls destroy(). That way, the Drop for S would be automatically generated.
    – rodrigo
    Nov 12 at 10:29






  • 1




    @rodrigo Isn't that exactly what the OP was trying to do?
    – Sven Marnach
    Nov 12 at 10:52










  • @SvenMarnach: Oh, I see. I was assuming S was some composite type. But then the NewType would be exactly like this S.
    – rodrigo
    Nov 12 at 11:02












  • 2




    Looks like you're not the only one to find this frustrating: github.com/gfx-rs/gfx/issues/2452
    – loganfsmyth
    Nov 12 at 2:07










  • Couldn't you use a NewType for T that implements Drop and calls destroy(). That way, the Drop for S would be automatically generated.
    – rodrigo
    Nov 12 at 10:29






  • 1




    @rodrigo Isn't that exactly what the OP was trying to do?
    – Sven Marnach
    Nov 12 at 10:52










  • @SvenMarnach: Oh, I see. I was assuming S was some composite type. But then the NewType would be exactly like this S.
    – rodrigo
    Nov 12 at 11:02







2




2




Looks like you're not the only one to find this frustrating: github.com/gfx-rs/gfx/issues/2452
– loganfsmyth
Nov 12 at 2:07




Looks like you're not the only one to find this frustrating: github.com/gfx-rs/gfx/issues/2452
– loganfsmyth
Nov 12 at 2:07












Couldn't you use a NewType for T that implements Drop and calls destroy(). That way, the Drop for S would be automatically generated.
– rodrigo
Nov 12 at 10:29




Couldn't you use a NewType for T that implements Drop and calls destroy(). That way, the Drop for S would be automatically generated.
– rodrigo
Nov 12 at 10:29




1




1




@rodrigo Isn't that exactly what the OP was trying to do?
– Sven Marnach
Nov 12 at 10:52




@rodrigo Isn't that exactly what the OP was trying to do?
– Sven Marnach
Nov 12 at 10:52












@SvenMarnach: Oh, I see. I was assuming S was some composite type. But then the NewType would be exactly like this S.
– rodrigo
Nov 12 at 11:02




@SvenMarnach: Oh, I see. I was assuming S was some composite type. But then the NewType would be exactly like this S.
– rodrigo
Nov 12 at 11:02












1 Answer
1






active

oldest

votes

















up vote
6
down vote



accepted










The safest, easiest way to do this is to use an Option:



struct T;

impl Drop for T
fn drop(&mut self)
println!("dropping T");



struct S
member: Option<T>,


impl Drop for S
fn drop(&mut self)
if let Some(t) = self.member.take()
destroy_t(t);




fn destroy_t(_t: T)
println!("destroy T");


fn main()
let _x = S member: Some(T) ;




You could choose to use unsafe code with ManuallyDrop and swap out the current value for an uninitialized one1:



use std::mem::self, ManuallyDrop;

struct T;

impl Drop for T
fn drop(&mut self)
println!("dropping T");



struct S
member: ManuallyDrop<T>,


impl Drop for S
fn drop(&mut self)
unsafe
let valid_t = mem::replace(&mut *self.member, mem::uninitialized());
destroy_t(valid_t);
// do *not* call ManuallyDrop::drop
;



fn destroy_t(_t: T)
println!("destroy T");


fn main()
let _x = S
member: ManuallyDrop::new(T),
;



1 Using mem::uninitialized is extremely dangerous and hard to get right, especially in generic contexts. Using the nightly MaybeUninit, this might look like



#![feature(maybe_uninit)]

use std::mem::self, ManuallyDrop, MaybeUninit;

struct T;

impl Drop for T
fn drop(&mut self)
println!("dropping T");



struct S
member: ManuallyDrop<MaybeUninit<T>>,


impl Drop for S
fn drop(&mut self)
let invalid_t = MaybeUninit::uninitialized();
let valid_t = mem::replace(&mut *self.member, invalid_t);
let valid_t = unsafe valid_t.into_inner() ;
destroy_t(valid_t);
// do *not* call ManuallyDrop::drop



fn destroy_t(_t: T)
println!("destroy T");


fn main()
let _x = S
member: ManuallyDrop::new(MaybeUninit::new(T)),
;



See also:



  • How to move one field out of a struct that implements Drop trait?





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    1 Answer
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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    6
    down vote



    accepted










    The safest, easiest way to do this is to use an Option:



    struct T;

    impl Drop for T
    fn drop(&mut self)
    println!("dropping T");



    struct S
    member: Option<T>,


    impl Drop for S
    fn drop(&mut self)
    if let Some(t) = self.member.take()
    destroy_t(t);




    fn destroy_t(_t: T)
    println!("destroy T");


    fn main()
    let _x = S member: Some(T) ;




    You could choose to use unsafe code with ManuallyDrop and swap out the current value for an uninitialized one1:



    use std::mem::self, ManuallyDrop;

    struct T;

    impl Drop for T
    fn drop(&mut self)
    println!("dropping T");



    struct S
    member: ManuallyDrop<T>,


    impl Drop for S
    fn drop(&mut self)
    unsafe
    let valid_t = mem::replace(&mut *self.member, mem::uninitialized());
    destroy_t(valid_t);
    // do *not* call ManuallyDrop::drop
    ;



    fn destroy_t(_t: T)
    println!("destroy T");


    fn main()
    let _x = S
    member: ManuallyDrop::new(T),
    ;



    1 Using mem::uninitialized is extremely dangerous and hard to get right, especially in generic contexts. Using the nightly MaybeUninit, this might look like



    #![feature(maybe_uninit)]

    use std::mem::self, ManuallyDrop, MaybeUninit;

    struct T;

    impl Drop for T
    fn drop(&mut self)
    println!("dropping T");



    struct S
    member: ManuallyDrop<MaybeUninit<T>>,


    impl Drop for S
    fn drop(&mut self)
    let invalid_t = MaybeUninit::uninitialized();
    let valid_t = mem::replace(&mut *self.member, invalid_t);
    let valid_t = unsafe valid_t.into_inner() ;
    destroy_t(valid_t);
    // do *not* call ManuallyDrop::drop



    fn destroy_t(_t: T)
    println!("destroy T");


    fn main()
    let _x = S
    member: ManuallyDrop::new(MaybeUninit::new(T)),
    ;



    See also:



    • How to move one field out of a struct that implements Drop trait?





    share|improve this answer


























      up vote
      6
      down vote



      accepted










      The safest, easiest way to do this is to use an Option:



      struct T;

      impl Drop for T
      fn drop(&mut self)
      println!("dropping T");



      struct S
      member: Option<T>,


      impl Drop for S
      fn drop(&mut self)
      if let Some(t) = self.member.take()
      destroy_t(t);




      fn destroy_t(_t: T)
      println!("destroy T");


      fn main()
      let _x = S member: Some(T) ;




      You could choose to use unsafe code with ManuallyDrop and swap out the current value for an uninitialized one1:



      use std::mem::self, ManuallyDrop;

      struct T;

      impl Drop for T
      fn drop(&mut self)
      println!("dropping T");



      struct S
      member: ManuallyDrop<T>,


      impl Drop for S
      fn drop(&mut self)
      unsafe
      let valid_t = mem::replace(&mut *self.member, mem::uninitialized());
      destroy_t(valid_t);
      // do *not* call ManuallyDrop::drop
      ;



      fn destroy_t(_t: T)
      println!("destroy T");


      fn main()
      let _x = S
      member: ManuallyDrop::new(T),
      ;



      1 Using mem::uninitialized is extremely dangerous and hard to get right, especially in generic contexts. Using the nightly MaybeUninit, this might look like



      #![feature(maybe_uninit)]

      use std::mem::self, ManuallyDrop, MaybeUninit;

      struct T;

      impl Drop for T
      fn drop(&mut self)
      println!("dropping T");



      struct S
      member: ManuallyDrop<MaybeUninit<T>>,


      impl Drop for S
      fn drop(&mut self)
      let invalid_t = MaybeUninit::uninitialized();
      let valid_t = mem::replace(&mut *self.member, invalid_t);
      let valid_t = unsafe valid_t.into_inner() ;
      destroy_t(valid_t);
      // do *not* call ManuallyDrop::drop



      fn destroy_t(_t: T)
      println!("destroy T");


      fn main()
      let _x = S
      member: ManuallyDrop::new(MaybeUninit::new(T)),
      ;



      See also:



      • How to move one field out of a struct that implements Drop trait?





      share|improve this answer
























        up vote
        6
        down vote



        accepted







        up vote
        6
        down vote



        accepted






        The safest, easiest way to do this is to use an Option:



        struct T;

        impl Drop for T
        fn drop(&mut self)
        println!("dropping T");



        struct S
        member: Option<T>,


        impl Drop for S
        fn drop(&mut self)
        if let Some(t) = self.member.take()
        destroy_t(t);




        fn destroy_t(_t: T)
        println!("destroy T");


        fn main()
        let _x = S member: Some(T) ;




        You could choose to use unsafe code with ManuallyDrop and swap out the current value for an uninitialized one1:



        use std::mem::self, ManuallyDrop;

        struct T;

        impl Drop for T
        fn drop(&mut self)
        println!("dropping T");



        struct S
        member: ManuallyDrop<T>,


        impl Drop for S
        fn drop(&mut self)
        unsafe
        let valid_t = mem::replace(&mut *self.member, mem::uninitialized());
        destroy_t(valid_t);
        // do *not* call ManuallyDrop::drop
        ;



        fn destroy_t(_t: T)
        println!("destroy T");


        fn main()
        let _x = S
        member: ManuallyDrop::new(T),
        ;



        1 Using mem::uninitialized is extremely dangerous and hard to get right, especially in generic contexts. Using the nightly MaybeUninit, this might look like



        #![feature(maybe_uninit)]

        use std::mem::self, ManuallyDrop, MaybeUninit;

        struct T;

        impl Drop for T
        fn drop(&mut self)
        println!("dropping T");



        struct S
        member: ManuallyDrop<MaybeUninit<T>>,


        impl Drop for S
        fn drop(&mut self)
        let invalid_t = MaybeUninit::uninitialized();
        let valid_t = mem::replace(&mut *self.member, invalid_t);
        let valid_t = unsafe valid_t.into_inner() ;
        destroy_t(valid_t);
        // do *not* call ManuallyDrop::drop



        fn destroy_t(_t: T)
        println!("destroy T");


        fn main()
        let _x = S
        member: ManuallyDrop::new(MaybeUninit::new(T)),
        ;



        See also:



        • How to move one field out of a struct that implements Drop trait?





        share|improve this answer














        The safest, easiest way to do this is to use an Option:



        struct T;

        impl Drop for T
        fn drop(&mut self)
        println!("dropping T");



        struct S
        member: Option<T>,


        impl Drop for S
        fn drop(&mut self)
        if let Some(t) = self.member.take()
        destroy_t(t);




        fn destroy_t(_t: T)
        println!("destroy T");


        fn main()
        let _x = S member: Some(T) ;




        You could choose to use unsafe code with ManuallyDrop and swap out the current value for an uninitialized one1:



        use std::mem::self, ManuallyDrop;

        struct T;

        impl Drop for T
        fn drop(&mut self)
        println!("dropping T");



        struct S
        member: ManuallyDrop<T>,


        impl Drop for S
        fn drop(&mut self)
        unsafe
        let valid_t = mem::replace(&mut *self.member, mem::uninitialized());
        destroy_t(valid_t);
        // do *not* call ManuallyDrop::drop
        ;



        fn destroy_t(_t: T)
        println!("destroy T");


        fn main()
        let _x = S
        member: ManuallyDrop::new(T),
        ;



        1 Using mem::uninitialized is extremely dangerous and hard to get right, especially in generic contexts. Using the nightly MaybeUninit, this might look like



        #![feature(maybe_uninit)]

        use std::mem::self, ManuallyDrop, MaybeUninit;

        struct T;

        impl Drop for T
        fn drop(&mut self)
        println!("dropping T");



        struct S
        member: ManuallyDrop<MaybeUninit<T>>,


        impl Drop for S
        fn drop(&mut self)
        let invalid_t = MaybeUninit::uninitialized();
        let valid_t = mem::replace(&mut *self.member, invalid_t);
        let valid_t = unsafe valid_t.into_inner() ;
        destroy_t(valid_t);
        // do *not* call ManuallyDrop::drop



        fn destroy_t(_t: T)
        println!("destroy T");


        fn main()
        let _x = S
        member: ManuallyDrop::new(MaybeUninit::new(T)),
        ;



        See also:



        • How to move one field out of a struct that implements Drop trait?






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Nov 12 at 2:13

























        answered Nov 12 at 2:01









        Shepmaster

        146k11281413




        146k11281413



























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