Question

使用函数指针在对象上调用方法的正确语法是什么？

struct Foo {
    var: i32,
}

impl Foo {
    fn method(&mut self, arg: i32) {
        self.var = self.var + arg;
        println!("var = {}", self.var);
    }
}

fn main() {
    let foo = Foo { var: 11 };
    let func_ptr: Fn() = &foo.method;
    (func_ptr).method(12);
}

我收到此错误：

error[E0615]: attempted to take value of method `method` on type `Foo`
  --> src/main.rs:14:31
   |
14 |     let func_ptr: Fn() = &foo.method;
   |                               ^^^^^^ help: use parentheses to call the method: `method(...)`

error[E0599]: no method named `method` found for type `dyn std::ops::Fn()` in the current scope
  --> src/main.rs:15:16
   |
15 |     (func_ptr).method(12);
   |                ^^^^^^
   |
   = note: (func_ptr) is a function, perhaps you wish to call it

error[E0277]: the size for values of type `dyn std::ops::Fn()` cannot be known at compilation time
  --> src/main.rs:14:9
   |
14 |     let func_ptr: Fn() = &foo.method;
   |         ^^^^^^^^ doesn't have a size known at compile-time
   |
   = help: the trait `std::marker::Sized` is not implemented for `dyn std::ops::Fn()`
   = note: to learn more, visit <https://doc.rust-lang.org/book/ch19-04-advanced-types.html#dynamically-sized-types-and-the-sized-trait>
   = note: all local variables must have a statically known size
   = help: unsized locals are gated as an unstable feature

我猜我没有为func_ptr类型使用正确的类型；什么是正确的类型？

Answer 1

方法不绑定self参数，但是您可以通过其完全限定的名称访问方法，并且可以根据需要将它们保存在函数指针中。稍后，当您调用函数指针时，您必须自己提供self参数：

struct Foo {
    var: i32,
}

impl Foo {
    fn method(&mut self, value: i32) {
        self.var += value;
        println!("var = {}", self.var);
    }
}

fn main() {
    let mut foo = Foo { var: 11 };
    let func_ptr = Foo::method;
    func_ptr(&mut foo, 12);
}

如果您想手动定义此局部变量的类型，则它将是以下内容：

let func_ptr: fn(&mut Foo, i32) = Foo::method;

如果要使用特征符号，则必须在参考后面使用它：

let func_ptr: &Fn(&mut Foo, i32) = &Foo::method;

如何通过函数指针调用方法？

1 个答案: