英文:
Is it possible to "advance" a mutable slice through a mutable reference?
问题
我经常用共享切片做这样的事情：
今天我注意到，当切片是可变的时，这种方法不起作用：
第二个示例会产生以下错误：
这是一个借用检查器的限制吗，还是我尝试做的事情实际上是不安全的？有没有一种解决方法？
英文:
I often do things like this with shared slices:
fn read_a_few_bytes(slice: &mut &[u8]) {
dbg!(&slice[..3]);
*slice = &slice[3..];
}
Today I noticed that that approach doesn't work when the slice is mutable:
fn read_a_few_bytes(slice: &mut &mut [u8]) {
dbg!(&slice[..3]);
*slice = &mut slice[3..];
}
The second example gives this error:
error: lifetime may not live long enough
--> src/main.rs:3:5
|
1 | fn read_a_few_bytes(slice: &mut &mut [u8]) {
| - - let's call the lifetime of this reference `'2`
| |
| let's call the lifetime of this reference `'1`
2 | dbg!(&slice[..3]);
3 | *slice = &mut slice[3..];
| ^^^^^^^^^^^^^^^^^^^^^^^^ assignment requires that `'1` must outlive `'2`
Is this a borrow checker limitation, or is there some way that what I'm trying to do here is actually unsound? Is there a workaround?
答案1
得分: 6
你可以使用 std::mem::take 来解决这个问题。
fn read_a_few_bytes(slice: &mut &mut [u8]) {
let data = std::mem::take(slice);
*slice = &mut data[3..];
}
英文:
You can use std::mem::take to solve this.
fn read_a_few_bytes(slice: &mut &mut [u8]) {
let data = std::mem::take(slice);
*slice = &mut data[3..];
}
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