函数可以接受参数并进行修改:
fn set_42(int: &mut i32) {
*int += 42;
}
fn main() {
let mut int = 0;
set_42(&mut int);
println!("{:?}", int);
}
输出:
42
将代码更改为使用切片会出现大量错误:
fn pop_front(slice: &mut [i32]) {
*slice = &{slice}[1..];
}
fn main() {
let mut slice = &[0, 1, 2, 3][..];
pop_front(&mut slice);
println!("{:?}", slice);
}
输出:
error[E0308]: mismatched types
--> src/main.rs:2:14
|
2 | *slice = &{ slice }[1..];
| ^^^^^^^^^^^^^^^
| |
| expected slice `[i32]`, found `&[i32]`
| help: consider removing the borrow: `{ slice }[1..]`
error[E0277]: the size for values of type `[i32]` cannot be known at compilation time
--> src/main.rs:2:5
|
2 | *slice = &{ slice }[1..];
| ^^^^^^ doesn't have a size known at compile-time
|
= help: the trait `std::marker::Sized` is not implemented for `[i32]`
= note: to learn more, visit <https://doc.rust-lang.org/book/ch19-04-advanced-types.html#dynamically-sized-types-and-the-sized-trait>
= note: the left-hand-side of an assignment must have a statically known size
如果我们尝试使用一个可变的切片(这并不是我真正想要的;我不想修改切片内部的值,我只想修改切片本身以覆盖更小范围的元素),和一个可变参数,它对原始切片没有影响:
fn pop_front(mut slice: &mut [i32]) {
slice = &mut {slice}[1..];
}
fn main() {
let mut slice = &mut [0, 1, 2, 3][..];
pop_front(&mut slice);
println!("{:?}", slice);
}
输出:
[0, 1, 2, 3]
有没有办法修改作为函数参数的切片?我不想修改切片中的元素;我只想修改切片本身的范围,使其成为一个更小的“子切片”。
pop_front
接受一个不可变 切片 的可变 引用。我的pop_front
接受一个可变切片的可变引用。 - Francis Gagné