经典的Builder Pattern要求在要构建的类中声明字段,并在构建器类中声明完全相同的字段。当存在许多字段时,这会导致问题,并且在重构期间,字段类型不会保持同步。这是我的意思的一个例子(我从Joshua Block的一篇文章中借用了这个代码示例):
// Builder Pattern
public class NutritionFacts {
private final int servingSize;
private final int servings;
private final int calories;
private final int fat;
private final int sodium;
private final int carbohydrate;
public static class Builder {
private int servingSize = 0;
private int servings = 0;
private int calories = 0;
private int fat = 0;
private int carbohydrate = 0;
private int sodium = 0;
public Builder servingSize(int val)
{ servingSize = val; return this; }
public Builder servings(int val)
{ servings = val; return this; }
public Builder calories(int val)
{ calories = val; return this; }
public Builder fat(int val)
{ fat = val; return this; }
public Builder carbohydrate(int val)
{ carbohydrate = val; return this; }
public Builder sodium(int val)
{ sodium = val; return this; }
public NutritionFacts build() {
return new NutritionFacts(this);
}
}
private NutritionFacts(Builder builder) {
servingSize = builder.servingSize;
servings = builder.servings;
calories = builder.calories;
fat = builder.fat;
sodium = builder.sodium;
carbohydrate = builder.carbohydrate;
}
}
现在让我们假设servingSize
需要从int
更改为long
并且此更改是在NutritonFacts
中完成的,但是,不小心,也在静态Builder
中完成。
不可否认,字段数量较少。 NutritionFacts
有6个字段,因此Builder
也是如此。如果有20或100个字段怎么办?在NutritionFacts
和Builder
中复制它们将是真正的痛苦。有没有更好的方法可以避免所有重复和类型同步错误的可能性?
答案 0 :(得分:3)
您可以使用NutritionFacts
对象存储构建器的状态:
// Builder Pattern
public class NutritionFacts {
private final int servingSize;
private final int servings;
private final int calories;
private final int fat;
private final int sodium;
private final int carbohydrate;
public static class Builder {
private NutritionFacts state = new NutritionFacts(0,0,0,0,0,0);
public Builder servingSize(int val) {
state = new NutritionFacts(val, state.servings, state.calories, state.fat, state.sodium, state.carbohydrate);
return this;
}
[...]
public NutritionFacts build() {
return new NutritionFacts(this);
}
}
private NutritionFacts(Builder builder) {
servingSize = builder.state.servingSize;
servings = builder.state.servings;
calories = builder.state.calories;
fat = builder.state.fat;
sodium = builder.state.sodium;
carbohydrate = builder.state.carbohydrate;
}
private NutritionFacts(int servingSize, int servings, int calories, int fat, int sodium, int carbohydrate) {
this.servingSize = servingSize;
this.servings = servings;
this.calories = calories;
this.fat = fat;
this.sodium = sodium;
this.carbohydrate = carbohydrate;
}
}
由于NutritionFacts
是不可变的,因此需要为每次更改构建一个新的状态对象,这可能是也可能不值得。
如果您可以使NutritionFacts
的内部状态相互关联,但使用私有设置器会更容易 - 根据定义而不是final
关键字使对象不可变:
// Builder Pattern
public class NutritionFacts {
private int servingSize = 0;
private int servings = 0;
private int calories = 0;
private int fat = 0;
private int sodium = 0;
private int carbohydrate = 0;
public static class Builder {
private NutritionFacts state = new NutritionFacts();
public Builder servingSize(int val) {
state.servingSize = val;
return this;
}
public Builder servings(int val) {
state.servings = val;
return this;
}
[...]
public NutritionFacts build() {
return new NutritionFacts(this);
}
}
private NutritionFacts(Builder builder) {
servingSize = builder.state.servingSize;
servings = builder.state.servings;
calories = builder.state.calories;
fat = builder.state.fat;
sodium = builder.state.sodium;
carbohydrate = builder.state.carbohydrate;
}
}
答案 1 :(得分:2)
您可以使用步骤构建器模式增强经典构建器模式,以便构建具有无脑界面,易于使用,不可能出错的对象。有关详细信息,请参阅此post。