Why is calling a FnOnce closure a move?
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
add a comment |
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
add a comment |
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
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
rust closures move-semantics
edited Nov 9 at 22:08
Shepmaster
147k11281415
147k11281415
asked Nov 9 at 22:01
user96425
382
382
add a comment |
add a comment |
1 Answer
1
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oldest
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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.
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
add a comment |
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.
add a comment |
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.
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.
edited Nov 9 at 22:39
answered Nov 9 at 22:09
Shepmaster
147k11281415
147k11281415
add a comment |
add a comment |
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