匹配實現特質的類型，因此返回它

Question

我正在嘗試一些特定的生成器模式。請不要介意這裏的設計 - 這是一個試用。

我有一個Renderer類型，它產生一個方法set_shape。給這個調用的參數類型應該實現空的特徵IsRenderable。 方法的登錄名應該在結構類型Rectangle和Circle之間有所不同，它們都實現了IsRenderable特徵。不要擔心返回類型。

#[derive(Clone, Copy)] 
pub struct Rectangle { 
    pub origin: Point, 
    pub height: usize, 
    pub width: usize, 
} 

trait IsRenderable {} 

impl IsRenderBuilder for Rectangle {} 

impl<'a> Renderer<'a> { 
    // logic that needs lifetime 'a 
    pub fn set_shape<T: IsRenderable>(shape: T) -> Box<IsRenderBuilder> { 
     match shape { 
      _ => panic!("WTF!"), 
     } 
    } 
} 

我想做這樣的事情。 僞：

match shape { 
    Rectangle => return RectangleShapeBuilder, 
    Circle => return CircleShapeBuilder, 
    _ => panic!("WTF!"), 
} 

Answer 1

而不是使用match的，你應該在你的IsRenderable特點，用於返回特定的實現者適當的參數中加入的方法。

#[derive(Clone, Copy)] 
pub struct Rectangle; 

pub trait IsRenderable { 
    fn new_builder(&self) -> Box<IsRenderBuilder>; 
} 

impl IsRenderable for Rectangle { 
    fn new_builder(&self) -> Box<IsRenderBuilder> { 
     Box::new(RectangleShapeBuilder) 
    } 
} 

struct RectangleShapeBuilder; 

pub trait IsRenderBuilder {} 

impl IsRenderBuilder for RectangleShapeBuilder {} 

pub fn set_shape<T: IsRenderable>(shape: T) -> Box<IsRenderBuilder> { 
    shape.new_builder() 
} 

你甚至可能使用相應的類型，以避免拳擊IsRenderBuilder：

#[derive(Clone, Copy)] 
pub struct Rectangle; 

pub trait IsRenderable { 
    type Builder: IsRenderBuilder; 

    fn new_builder(&self) -> Self::Builder; 
} 

impl IsRenderable for Rectangle { 
    type Builder = RectangleShapeBuilder; 

    fn new_builder(&self) -> Self::Builder { 
     RectangleShapeBuilder 
    } 
} 

pub struct RectangleShapeBuilder; 

pub trait IsRenderBuilder {} 

impl IsRenderBuilder for RectangleShapeBuilder {} 

pub fn set_shape<T: IsRenderable>(shape: T) -> T::Builder { 
    shape.new_builder() 
}