rust - 如何将通用结构作为函数参数

Question

我试图编写一个将通用结构作为参数的函数:

struct S<T> {
    v: T,
}

fn foo(a: &S) {
    a.v
}

但是会发生错误:

error[E0107]: wrong number of type arguments: expected 1, found 0
 --> src/main.rs:5:12
  |
5 | fn foo(a: &S) {
  |            ^ expected 1 type argument

error: aborting due to previous error

第一次更新
更改为

struct S<T> {
    v: T,
}

fn foo<T>(a: &S<T>) -> T {
    a.v
}

发生另一个错误:

error[E0507]: cannot move out of `a.v` which is behind a shared reference
 --> src/main.rs:6:5
  |
6 |     a.v
  |     ^^^ move occurs because `a.v` has type `T`, which does not implement the `Copy` trait

error: aborting due to previous error

第二次更新
终于，我得到了我想要的:

struct S<T> {
    v: T,
}

fn foo<T: Copy>(a: &S<T>) -> T {
    a.v
}

Answer 1

您错过了通用参数

struct S<T> {
    v: T,
}

fn foo<T>(a: S<T>) -> T {
    a.v
}

Or use the Copy bound

fn foo<T: Copy>(a: &S<T>) -> T {
  a.v
}