Question

在很多情况下，我都有一个很长的布尔表达式来求值：

fun clickClickableSpan(textToClick: CharSequence): ViewAction {
    return object : ViewAction {

        override fun getConstraints(): Matcher<View> {
            return Matchers.instanceOf(TextView::class.java)
        }

        override fun getDescription(): String {
            return "clicking on a ClickableSpan";
        }

        override fun perform(uiController: UiController, view: View) {
            val textView = view as TextView
            val spannableString = textView.text as SpannableString

            if (spannableString.isEmpty()) {
                // TextView is empty, nothing to do
                throw NoMatchingViewException.Builder()
                        .includeViewHierarchy(true)
                        .withRootView(textView)
                        .build();
            }

            // Get the links inside the TextView and check if we find textToClick
            val spans = spannableString.getSpans(0, spannableString.length, ClickableSpan::class.java)
            if (spans.isNotEmpty()) {
                var spanCandidate: ClickableSpan
                for (span: ClickableSpan in spans) {
                    spanCandidate = span
                    val start = spannableString.getSpanStart(spanCandidate)
                    val end = spannableString.getSpanEnd(spanCandidate)
                    val sequence = spannableString.subSequence(start, end)
                    if (textToClick.toString().equals(sequence.toString())) {
                        span.onClick(textView)
                        return;
                    }
                }
            }

            // textToClick not found in TextView
            throw NoMatchingViewException.Builder()
                    .includeViewHierarchy(true)
                    .withRootView(textView)
                    .build()

        }
    }
}

这应该返回一个布尔值！

a，b，c，d，e和f是简单的相等比较，并且都不同。这持续了12行，给我带来了44的圈复杂度

我试图查看Map对象以降低复杂性，但没有找到解决方法。

如何减少这种表达式的复杂性？

Answer 1

您无法降低给定表达式的复杂度。

一种较慢的方法（但可能更结构化）是对表达式进行分组，并对外部数组使用Array#some进行评估，对内部数组使用Array#every进行评估。

return [[a, b, c], [d, e, f]].some(a => a.every(Boolean));

如何减少长布尔表达式的复杂度？

1 个答案: