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，如何以正确的方式实施？

Answer 1

我宁愿使用Java 8以来可用的Arrays和Streams。例如：

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

如果您需要一小部分元素的更好的性能解决方案，那么在内存和进程效率上都是有效的，并且不使用语法糖或Java 8流（而循环更清晰，由于Vinod Singh Bist而得到改进）：

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 2

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

如果数组不是sorted，则结果为undefined。您应该Arrays.sort(validOperators);

Answer 3

请尝试以下提到的解决方案。

    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);

扫描仪和用户输入

3 个答案: