Java中的用户定义方法

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2 年前已关闭。

Answer 1

首先，我将删除功能中的println，并使其成为双重返回功能。

也许您要寻找的是定义简短方法的lambdas。
在这里，我们可以使用 doubleBinaryOperator 。

public static void main(String[] args) {       
    DoubleBinaryOperator add = (x, y) -> x + y;
    DoubleBinaryOperator sub = (x, y) -> x - y;
    DoubleBinaryOperator mul = (x, y) -> x * y;
    DoubleBinaryOperator div = (x, y) -> x / y;
    double num1 = 5;
    double num2 = 10;
    System.out.printf("sum %f%nsub %f%nmul %f%ndiv %f%n",
        add.applyAsDouble(num1, num2),
        sub.applyAsDouble(num1 + num2, num1),
        mul.applyAsDouble(num1, num2),
        div.applyAsDouble(num1 + num2, (num1 + num2) - num1));
}

不幸的是.applyasdouble有点尴尬。 applyasdouble应解释为： apply 两个（double）参数，并将结果作为double 值。

您的解决方案也很好。

Answer 2

public static void main(String[] args) {
    double num1 = 5;
    double num2 = 10;
    BiConsumer<Double, Double> add = (a, b) -> System.out.println("Sum is: " + (a + b));
    BiConsumer<Double, Double> sub = (a, b) -> System.out.println("Difference is: " + ((a + b) - a));
    BiConsumer<Double, Double> mul = (a, b) -> System.out.println("Multiplication is: " + (a * b));
    BiConsumer<Double, Double> div = (a, b) -> System.out.println("Division is: " + ((a + b) / ((a + b) - a)));
    Stream.of(add, sub, mul, div).forEach(x -> x.accept(num1, num2));
}

您可以这样写。 :)

Answer 3

    double num1 = 5;
    double num2 = 10;
    double sum=sum(num1,num2);
    double sub=sub(sum,num1);
    double mul=mul(num1,num2);
    double div=div(sum,sum-1);
    System.out.println("sum is: "+sum+"\n Difference is: "+sub+"\nMultiplication is: "+mul+"\nDivision is: "+div);
    sum(num1, num2);
    sub(num1 + num2, num1);
    mul(num1, num2);
    div(num1 + num2, (num1 + num2) - num1);
}

public static double sum(double num1, double num2)
{
    return num1+num2;
}

public static double sub(double sumVal, double num1)
{
    return sumVal-num1;
}

public static double mul(double num1, double num2)
{
    return num1*num2;
}

public static double div(double sumVal, double subVal)
{
    return sumVal/subVal;
}