关键就在上一句:
Rust 显式地允许使用不相交的结构字段来完成[重新借用到多个可变引用],因为不相交可以被静态证明
在这种情况之外,编译器无法判断两个借用是不相交的。实际上,这意味着编译器无法判断函数调用产生的借用将是不相交的。
struct Thing {
a: i32,
b: i32,
}
fn example_works(thing: &mut Thing) {
let a = &mut thing.a;
let b = &mut thing.b;
}
fn get_a(thing: &mut Thing) -> &mut i32 {
&mut thing.a
}
fn get_b(thing: &mut Thing) -> &mut i32 {
&mut thing.b
}
fn example_doesnt_work(thing: &mut Thing) {
let a = get_a(thing);
let b = get_b(thing);
println!("{}, {}", a, b);
}
error[E0499]: cannot borrow `*thing` as mutable more than once at a time
--> src/lib.rs:26:19
|
25 | let a = get_a(thing);
| ----- first mutable borrow occurs here
26 | let b = get_b(thing); // cannot borrow `*thing` as mutable more than once at a time
| ^^^^^ second mutable borrow occurs here
27 | println!("{}, {}", a, b);
| - first borrow later used here
这会发生在元组结构中吗?
没有具体说because它是一个元组结构,但是,是的,它可能出于相同的原因而发生。如果您从函数调用中获得借用,您将遇到与“传统”结构相同的问题。