我正在尝试用 BinaryOperator 功能接口(interface)替换算术运算的通用开关。

基本方法是:

private static int computeOne(int res, String operand, String operation) {
    int number = Integer.parseInt(operand);

    switch (operation) {
        case "+":
            res += number;
            break;
        case "-":
            res -= number;
            break;
        case "*":
            res *= number;
            break;
        case "/":
            res = (number != 0 ? res / number : Integer.MAX_VALUE);
            break;
        default:
            res = 0;
            System.out.println("unknown operation");
    }

    return res;
}

据我所知，写这样的东西是必要的:

BinaryOperator<Integer> action = (a,b) -> computeExpression(a + operation + b);
action.apply(res, operand);

但我不明白如何避免 computeExpression 中的 switch 与 computeOne 相同。

最佳答案

你可以定义一个BinaryOperator<Integer>对于每个算术运算:

// a = operand 1
// b = operand 2
(a, b) -> a * b;
(a, b) -> a + b;
(a, b) -> a / b;
(a, b) -> a - b;

然后你可以应用一个传递 2 个参数:

// result = operation.apply(a, b);
int result = ((BinaryOperator<Integer>) ((a, b) -> a * b)).apply(2, 2);

我会使用枚举来枚举这些操作:

class Test {

    public static void main(String[] args) {
         System.out.println(computeOne(4, "2", "/"));  // 2
         System.out.println(computeOne(4, "2", "*"));  // 8
         System.out.println(computeOne(4, "2", "-"));  // 2
         System.out.println(computeOne(4, "2", "+"));  // 6
    }

    private static int computeOne(int res, String operand, String operation) {
        return Operation.getOperationBySymbol(operation)
                        .getBinaryOperator()
                        .apply(res, Integer.parseInt(operand));
    }

    private enum Operation {
        // operation = symbol, action
        MULTIPLICATION("*", (a, b) -> a * b),
        ADDITION("+", (a, b) -> a + b),
        SUBTRACTION("-", (a, b) -> a - b),
        DIVISION("/", (a, b) -> a / b);

        private final BinaryOperator<Integer> binaryOperator;
        private final String symbol;

        Operation(String symbol, BinaryOperator<Integer> binaryOperator) {
            this.symbol = symbol;
            this.binaryOperator = binaryOperator;
        }

        public BinaryOperator<Integer> getBinaryOperator() {
            return binaryOperator;
        }

        public String getSymbol() {
            return symbol;
        }

        public static Operation getOperationBySymbol(String symbol) {
            for (Operation operation : values()) {
                if (operation.getSymbol().equals(symbol)) {
                    return operation;
                }
            }

            throw new IllegalArgumentException("Unknown symbol: " + symbol);
        }
    }

}

您也可以使用 BiFunction<BinaryOperator<?>, Pair<?, ?>, ?> 来“简化”它:

// BiFunction<Operator, Operands, Result>
// Operator = BinaryOperator<?>
// Operands = Pair<?, ?>
BiFunction<BinaryOperator<Integer>, Pair<Integer, Integer>, Integer> f = 
    (operator, operands) -> 
        operator.apply(operands.getKey(), operands.getValue());

f.apply((a, b) -> a + b, new Pair<>(2, 2)); // 4

关于java - 用 BinaryOperator 替换 switch，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/49699781/