java 表达式太复杂减少条件运算符的数量

Question

我们有一个程序根据我们的代码运行以遵守一些编码标准。

程序说：

Expression should not be too complex, reduce the number of conditional operators used int he expression Min allowed 3.

如何减少条件运算符的数量？也许将关键事件放在一个数组中？

public boolean onlyNumbers(KeyEvent evt) {
    char c = evt.getKeyChar();
    boolean returnValue = true;
    if (
        !(
            Character.isDigit(c) 
            || c == KeyEvent.VK_BACK_SPACE
            || c == KeyEvent.VK_DELETE 
            || c == KeyEvent.VK_END 
            || c == KeyEvent.VK_HOME
        )
        || c == KeyEvent.VK_PAGE_UP
        || c == KeyEvent.VK_PAGE_DOWN
        || c == KeyEvent.VK_INSERT
    ) {
        evt.consume();
        returnValue = false;
    }
    return returnValue;
}

Answer 1

您可以这样做的一种方法是构造一个 HashSet<Character> keys 包含您要测试的所有字符，然后使用 keys.contains(c) 来测试它是否是其中之一。

或者，您可以使用 switch 语句，让所有这些字符落入同一代码块。

但我个人最喜欢忽略警告。代码非常清楚（modulo ajb 关于括号的评论）。

Answer 2

这是一种方法。可以通过使用数组来改进。

public boolean onlyNumbers(KeyEvent evt) {
    char c = evt.getKeyChar();
    boolean returnValue;

    returnValue =  !(Character.isDigit(c));
    returnValue &= !(c == KeyEvent.VK_BACK_SPACE);
    returnValue &= !(c == KeyEvent.VK_DELETE);
    returnValue &= !(c == KeyEvent.VK_END);
    returnValue &= !(c == KeyEvent.VK_HOME);
    returnValue |= (c == KeyEvent.VK_PAGE_UP);
    returnValue |= (c == KeyEvent.VK_PAGE_DOWN);
    returnValue |= (c == KeyEvent.VK_INSERT);
    if(returnValue)
    {
        evt.consume();
        returnValue = !returnValue;
    }
    return returnValue;
}

前五个和最后三个作业可以在各自的分组中压缩为两个总作业，但这将归结为您的编码标准。

Answer 3

final String junkChars = new String(new char[] {
    KeyEvent.VK_BACK_SPACE,
    KeyEvent.VK_DELETE,
    KeyEvent.VK_END,
    KeyEvent.VK_HOME
    /* The last three seem unused in the latest version
    KeyEvent.VK_PAGE_UP,
    KeyEvent.VK_PAGE_DOWN,
    KeyEvent.VK_INSERT */
});
if(!Character.isDigit(c) && junkChars.indexOf(c) == -1) {
   evt.consume();
   return false;
}  else {
    return true;
}

Answer 4

您的示例的严格重构如下所示：

public boolean onlyNumbers(KeyEvent evt) {
    char c = evt.getKeyChar();
    boolean returnValue = true;
    boolean bad = Character.isDigit(c);
    bad |= (c == KeyEvent.VK_BACK_SPACE);
    bad |= (c == KeyEvent.VK_DELETE);
    bad |= (c == KeyEvent.VK_END);
    bad |= (c == KeyEvent.VK_HOME);
    boolean good = (c == KeyEvent.VK_PAGE_UP);
    good |= c == KeyEvent.VK_PAGE_DOWN;
    good |= c == KeyEvent.VK_INSERT;
    if (!bad || good) {
        evt.consume();
        returnValue = false;
    }
    return returnValue;
}

这凸显了其他人对您放置括号的关注

java 表达式太复杂减少条件运算符的数量

java Expression too complex reduce the number of conditionl operators

java

conditional-operator