Question

我是TypeScript的新手，请告诉我如何优化这段代码中的数据类型，这是我一直以来number不会写的。我了解到可以做到这一点，可以通过generics告诉我如何正确简要地描述此代码

interface ICalculator {
    add(num: number): number;
    subtract(num: number): number;
    multiply(num: number): number;
    divide(num: number): number;
    print(): void;
}

class Calculator implements ICalculator {
    protected _res: number;

    add(num:number) {
        return this._res = num + num;
    }

    subtract(num:number) {
        return this._res = num - num;
    }

    multiply(num:number) {
        return this._res = num * num;
    }

    divide(num:number) {
        return this._res = num / num;
    }

    print():void {
        console.log(`Currently value: ${this._res}`);
    }
}
let calc1 = new Calculator();
calc1.add(1);calc1.add(1);
calc1.print();

Answer 1

由于您已经写出了接口中每个函数的类型定义，因此在定义它们时可以跳过向其重新添加类型的过程。

这在类型上是相同的：

interface ICalculator {
    add(num: number): number;
    subtract(num: number): number;
    multiply(num: number): number;
    divide(num: number): number;
    print(): void;
}

class Calculator implements ICalculator {
    protected _res: number;

    add(num) {
        return this._res = num + num;
    }

    subtract(num) {
        return this._res = num - num;
    }

    multiply(num) {
        return this._res = num * num;
    }

    divide(num) {
        return this._res = num / num;
    }

    print() {
        console.log(`Currently value: ${this._res}`);
    }
}

您还可以定义函数的类型以进行更多重构：

type NumberOperation = (num: number) => number

interface ICalculator {
    add: NumberOperation
    subtract: NumberOperation
    multiply: NumberOperation
    divide: NumberOperation
    print(): void;
}

编辑： Titian Cernicova-Dragomir是正确的Class method parameter are never inferred, they will be implicitly typed to any

如果您不想在类内键入每个函数，则可以在声明类实例时使用该接口：

const calc: ICalculator = new Calculator()

calc.add('string') // type error

如何在此计算器中优化类型说明？

1 个答案: