Java 如何在 lambda 中烘焙值

发布于09月13日

假设我有这样的代码


@FunctionalInterface
interface Callable {
    Integer multiply(Integer by);
}

class Builder {
    private Integer multiplier;

    public Builder multiplyBy(Integer multiplier) {
        this.multiplier= multiplier;
        return this;
    }

    public Callable build() {
        return (value) -> value * multiplier; //how to "bake" hever value
    }
}

// calling

Builder builder;
Callable multiplierBy3 = builder.multiplyBy(3);
Callable multiplierBy6 = builder.multiply(6);

multiplierBy3.multiply(10); //should return 30, returns 60
multiplierBy6.multiply(10); // return 60, ok, propably because of new multipier value

之所以发生这种行为，是因为第二次调用multiplyBy会在Builder中设置新的value. 我如何才能创建每个lambda自己的乘数函数？

推荐答案

就像@Kayaman说的那样:

使用两个构建器，这就是您要维护lambda状态的地方.

因此，解决方案是这样的:

Builder builderBy3 = new Builder();
Builder builderBy6 = new Builder();

Callable multiplierBy3 = builderBy3.multiplyBy(3).build();
Callable multiplierBy6 = builderBy6.multiplyBy(6).build();

System.out.println(multiplierBy3.multiply(10)); // Returns 30
System.out.println(multiplierBy6.multiply(10)); // Returns 60

但您也可以修改Builder:

class Builder {
    private Integer multiplier;

    public Builder multiplyBy(Integer multiplier) {
        this.multiplier = multiplier;
        return this;
    }

    public Callable build() {
        int currentMultiplier = this.multiplier; // Current multiplier value
        return (value) -> value * currentMultiplier; // Use it in the lambda
    }
}

然后，您可以执行以下操作:

Builder builder = new Builder();

Callable multiplierBy3 = builder.multiplyBy(3).build();
Callable multiplierBy6 = builder.multiplyBy(6).build();

System.out.println(multiplierBy3.multiply(10)); // Returns 30
System.out.println(multiplierBy6.multiply(10)); // Returns 60