如何避免使用传递给“impl Trait”参数的值？

Question

在将值传递给以 impl Trait 作为参数的函数后，如何继续使用值的所有权？我已经尝试过将参数作为参考传递并且没有 & 但它们都不起作用。

trait Noise{
    fn make_noise(&self);
    fn call_by_name(&self);
}

struct Person {
    name: String,
}

impl Noise for Person{
    fn make_noise(&self){
        println!("Hello");
    }

    fn call_by_name(&self) {
        println!("Hey, {:?}, how are you doing?", self.name)
    }
}

fn talk(noisy: impl Noise){
    noisy.make_noise();
}

fn main() {
    let john_person= Person{name:String::from("John")};
    talk(john_person); 
    john_person.call_by_name(); // ERROR WHEN CALLING THE FUNCTION.
}

Answer 1

您应该使 fn talk 通过引用而不是通过值获取特征对象：

trait Noise {
    fn make_noise(&self);
    fn call_by_name(&self);
}

struct Person {
    name: String,
}

impl Noise for Person {
    fn make_noise(&self) {
        println!("Hello");
    }

    fn call_by_name(&self) {
        println!("Hey, {:?}, how are you doing?", self.name)
    }
}

fn talk(noisy: &impl Noise) {  // <- used &impl Noise
    noisy.make_noise();
}

fn main() {
    let john_person = Person {
        name: String::from("John"),
    };
    talk(&john_person);           // call by reference, not move
    john_person.call_by_name();
}

否则，john_person 将移至函数talk。这意味着您无法再访问它。

如果通过引用传递它，它将被借用直到函数结束。