扫描仪和用户输入

Question

使用下面的代码，当我使用+，-和Q时，循环不会结束。String[] validOperators = {"+", "-", "/", "*", "=", "q", "Q"};String userInput; Scanner scanner = new Scanner(System.in);System.out.print("Please enter an operation (+, -, /, *, = or Q to quit): ");userInput = scanner.nextLine();while(Arrays.binarySearch(validOperators, userInput) <= -1) {    System.out.print("Invalid input (+, -, /, *, = or Q to quit): ");    userInput = scanner.nextLine();}为什么会发生这种情况，我如何以正确的方式实施？

Answer 1

请尝试下面提到的解决方案。

    String[] validOperators = {"+", "-", "/", "*", "=", "q", "Q"};

    String userInput; 

    Scanner scanner = new Scanner(System.in);

    Arrays.sort(validOperators);

    do{

        System.out.print("Please enter a valid operation ( +, -, /, *, = , q or Q ) to quit: ");

        userInput = scanner.nextLine();

    }while(Arrays.binarySearch(validOperators, userInput) <= -1);

Answer 2

我宁愿使用数组和流，从Java 8开始就可用。例如：

Arrays.stream(validOperators).anyMatch(userInput::equals)

如果你需要一个更好的性能解决方案来满足一小部分元素，那么内存和进程都有效，并且不使用语法糖或Java 8流（而由于Vinod Singh Bist，循环更清晰，改进）：

public static void main(String[] args) {

    char[] validOperators = {'+', '-', '/', '*', '=', 'q', 'Q'}; // String is more expensive

    char userInput;

    Scanner scanner = new Scanner(System.in);

    do{

        System.out.print("Please enter a valid operation ( +, -, /, *, = , q or Q ) to quit: ");

        userInput = scanner.next().charAt(0);

    }while(!contains(validOperators, userInput)) ;

}

private static boolean contains(char[] elements, char c) {

    // for loop is usually faster for small lists than any built-in iterator for primitives like char

    for (int i = elements.length - 1; i >= 0; i--) {

        if (elements[i] == c) {

            return true;

        }

    }

    return false;

}

Answer 3

Arrays.binarySearch(validOperators, userInput)需要排序数组。

如果数组不是 ，则结果为 。你应该你sortedundefinedArrays.sort(validOperators);