How to "bake" value in lambda

Question

Let's say I have code like this

@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

Behavior happens because calling second time multiplyBy sets new value in builder. How I can manage to create each lambda own multiplier function?

Answer 1

Like @Kayaman said:

Use two builders, that's where you're keeping the lambda state.

So the solution will be this way:

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

But you can also do it modifying the 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
    }
}

Then you can do:

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