Closed
Description
Given the following code: https://play.rust-lang.org/?version=stable&mode=debug&edition=2021&gist=4e10570825edd99bc5f0f03fe82fa34f
fn foo(slice_a: &mut [u8], slice_b: &mut[u8]) {
core::mem::swap(&mut slice_a, &mut slice_b);
}
fn main() {
let a = [1u8,2,3];
let b = [4u8,5,6];
foo(&mut a, &mut b);
}The current output is:
error[E0623]: lifetime mismatch
--> src/main.rs:2:35
|
1 | fn foo(slice_a: &mut [u8], slice_b: &mut[u8]) {
| --------- -------- these two types are declared with different lifetimes...
2 | core::mem::swap(&mut slice_a, &mut slice_b);
| ^^^^^^^^^^^^ ...but data from `slice_b` flows into `slice_a` here
error[E0623]: lifetime mismatch
--> src/main.rs:2:35
|
1 | fn foo(slice_a: &mut [u8], slice_b: &mut[u8]) {
| --------- --------
| |
| these two types are declared with different lifetimes...
2 | core::mem::swap(&mut slice_a, &mut slice_b);
| ^^^^^^^^^^^^ ...but data from `slice_a` flows into `slice_b` here
For more information about this error, try `rustc --explain E0623`.
error: could not compile `playground` due to 2 previous errors
Ideally the output should look (something) like:
Hint: Each elided lifetime in input position becomes a distinct lifetime. To avoid this, explicitly declare a lifetime and assign it to both
Beginners likely aren't aware of lifetime elision rules, they may be thinking "but there are NO lifetimes here"...