如何在 return 类型中使用带有匿名闭包的高阶特征边界
How to use Higher Rank Trait Bounds with anonymous closure in return type
是否有可能 return FnMut
引用的闭包和 return 具有相同生命周期的引用?
fn fun(buf: &mut [f32], mut idx: usize) -> impl FnMut(&[i16]) -> &[i16] {
|input| {
buf[idx] = input[0] as f32;
idx += 1;
&input[1..]
}
}
我已经尝试过 impl for<'a> FnMut(&'a [i16]) -> &'a [i16])
之类的方法,它给出了
error[E0482]: lifetime of return value does not outlive the function call
--> src/main.rs:1:44
|
1 | fn fun(buf: &mut [f32], mut idx: usize) -> impl for<'a> FnMut(&'a [i16]) -> &'a [i16] {
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
|
note: the return value is only valid for the anonymous lifetime defined on the function body at 1:13
--> src/main.rs:1:13
|
1 | fn fun(buf: &mut [f32], mut idx: usize) -> impl for<'a> FnMut(&'a [i16]) -> &'a [i16] {
| ^^^^^^^^^^
- 返回的函数应该按值捕获
buf
(即使用 move
)
- 返回的函数不能超过
buf
(在下面的代码片段中具有生命周期 'buf
):
所以:
fn fun<'buf>(buf: &'buf mut [f32], mut idx: usize) -> impl FnMut(&[i16]) -> &[i16] + 'buf {
move |input| {
buf[idx] = input[0] as f32;
idx += 1;
&input[1..]
}
}
是否有可能 return FnMut
引用的闭包和 return 具有相同生命周期的引用?
fn fun(buf: &mut [f32], mut idx: usize) -> impl FnMut(&[i16]) -> &[i16] {
|input| {
buf[idx] = input[0] as f32;
idx += 1;
&input[1..]
}
}
我已经尝试过 impl for<'a> FnMut(&'a [i16]) -> &'a [i16])
之类的方法,它给出了
error[E0482]: lifetime of return value does not outlive the function call
--> src/main.rs:1:44
|
1 | fn fun(buf: &mut [f32], mut idx: usize) -> impl for<'a> FnMut(&'a [i16]) -> &'a [i16] {
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
|
note: the return value is only valid for the anonymous lifetime defined on the function body at 1:13
--> src/main.rs:1:13
|
1 | fn fun(buf: &mut [f32], mut idx: usize) -> impl for<'a> FnMut(&'a [i16]) -> &'a [i16] {
| ^^^^^^^^^^
- 返回的函数应该按值捕获
buf
(即使用move
) - 返回的函数不能超过
buf
(在下面的代码片段中具有生命周期'buf
):
所以:
fn fun<'buf>(buf: &'buf mut [f32], mut idx: usize) -> impl FnMut(&[i16]) -> &[i16] + 'buf {
move |input| {
buf[idx] = input[0] as f32;
idx += 1;
&input[1..]
}
}