Question

我获得了一个测试驱动开发任务的测试套件。该程序播放Rock，Paper，Scissors游戏，我还获得了实际程序的骨架代码。我不应该更改测试套件，而是我必须更改，或者在实际程序中生成代码以便测试通过。我在通过getInput方法时遇到了问题。这是实际程序的代码：

   public static char getInput(String prompt, char[] options, Scanner sc) {
 // getInput method
 // prompts user for an input that matches one of the given characters
 // if its not one of those, repeat. (use contains (above))
 char c;
 boolean flag = false;
 do {
   c = sc.next().charAt(0);
   if (contains(c, options)) {
     System.out.println(prompt + " ( y, n, q ):");
     flag = true;
   }
 }
 while (!flag);
return c; 
}

这是测试套件中测试getInput方法的方法：`

    private static void testGetInput() {
    OutputStream out;

out = resetSystemOut();
assert 'y' == RPS.getInput("Choose", new char[] {'y','n','q'}, new Scanner("y\n"));
assertOutput("Choose ( y, n, q ):\n", out);
out = resetSystemOut();
assert 'n' == RPS.getInput("Alice", new char[] {'y','n','q'}, new Scanner("n\n"));
assertOutput("Alice ( y, n, q ):\n", out);
out = resetSystemOut();
assert 'q' == RPS.getInput("Bob", new char[] {'y','n','q'}, new Scanner("q\n"));
assertOutput("Bob ( y, n, q ):\n", out);
out = resetSystemOut();
assert 'q' == RPS.getInput("Cloe", new char[] {'y','n','q'}, new Scanner("x\nw\nq\n")); //line 81
assertOutput("Cloe ( y, n, q ):\n" +
    "Cloe ( y, n, q ):\n" +
    "Cloe ( y, n, q ):\n", out);
out = resetSystemOut();
assert 'v' == RPS.getInput("Doug", new char[] {'v'}, new Scanner("vvvv\nv\n"));
assertOutput("Doug ( v ):\n" +
    "Doug ( v ):\n", out);
}

这是我得到的错误：

Exception in thread "main" java.lang.AssertionError: 54 18
Cloe ( y, n, q ):
Cloe ( y, n, q ):
Cloe ( y, n, q ):
Cloe ( y, n, q ):

        at RPSTester.assertOutput(RPSTester.java:226)
        at RPSTester.testGetInput(RPSTester.java:81)
        at RPSTester.main(RPSTester.java:25) //line 25 has a call to the testGetInput method

我在办公时间拜访了导师，但在揭示问题的全部答案之前，他只能告诉我。我会感谢任何帮助，暗示，指出我的错误;任何事情都表示赞赏。

Answer 1

你应该做这样的事情

 while(scanner.hasNext()) {
   String input = scanner.next();
   //check if input equals to y or n or q . If yes then break
   // otherwise System.out.println(prompt + " ( y, n, q ):")  
 }

Answer 2

您需要通过连接提示和选项中的字符来构建print语句，然后将该语句打印的次数与Scanner输入中的字符一样多。您可能必须调整新行字符\ n的追加以匹配您的测试断言。

这应该有所帮助：

public static char getInput(String prompt, char[] options, Scanner sc) {
        // getInput method
        // prompts user for an input that matches one of the given characters
        // if its not one of those, repeat. (use contains (above))
        char c;
        boolean flag = false;
        String printStr = prompt + " (";
        for (int i = 0; i < options.length; i++) {
            printStr += " " + options[i] + ",";
        }
        printStr = printStr.substring(0, printStr.length() - 1) + " ):\\n";

        do {
            c = sc.next().charAt(0);
            System.out.println(printStr);
            if (contains(c, options)) {
                flag = true;
            }
        }
        while (!flag);
        return c;
    }

测试驱动开发，Java，遇到断言错误

2 个答案: