Question

eclipse中有metrics plugin通过计算决策（独立路径）来衡量cyclomatic complexity，具体来说：

 (if, for, while, do, case, catch and the ?: ternary operator, as well as the && and || conditional logic operators in expressions)

例如，这里有一个得分为14的方法（每个决策路径1 +）

  String getMonthName (int month) {
        switch (month) {
                case 0: return "January";
                case 1: return "February";
                case 2: return "March";
                case 3: return "April";
                case 4: return "May";
                case 5: return "June";
                case 6: return "July";
                case 7: return "August";
                case 8: return "September";
                case 9: return "October";
                case 10: return "November";
                case 11: return "December";
                default: throw new IllegalArgumentException();
        }
}

我想知道在没有java中提到的分支的情况下是否有办法做出决定，这会破坏评估。基本上，我希望在没有插件检测它们的情况下做出不同的决定。指标显示出比实际情况更低的圈复杂度（独立路径/决策）。

Answer 1

为了降低圈复杂度，您应该解耦代码或重构代码以实现更好的实现。

对于提供的示例，您可以使用数组：

static final String[] MONTHS = {"January", "February", ... };
static String getMonthName(int month) {
    if (month < 0 || month >= MONTHS.length) {
        throw new IllegalArgumentException(String.format("Month %d doesn't exist", month));
    }
    return MONTHS[month];
}

即使是上述方法也可以解耦：

static final String[] MONTHS = {"January", "February", ... };
static String getMonthName(int month) {
    checkBoundaries(month, 0, MONTHS.length - 1, String.format("Month %d doesn't exist", month));
    return MONTHS[month];
}
static void checkBoundaries(int n, int lowerBound, int upperBound, String errorMessage) {
    if (n < lowerBound || n > upperBound) {
        throw new IllegalArgumentException(errorMessage);
    }
}

如何减少一个返回月份名称的方法的圈复杂度？

1 个答案: