java - 如何通过重构代码来降低环复杂度？

Question

这个问题重复了一个，但我仍然在问，因为通过使用解决方案中建议的方法，我无法显着降低复杂性。 函数复杂度是 28，我必须将其降低到 10 以下。

private void adjustViewport(IEditorPart editorPart, LineRange range,
    TextSelection selection) {
    ITextViewer viewer = EditorAPI.getViewer(editorPart);
    if (viewer == null) +1
        return; +1

    IDocument document = viewer.getDocument();
    LineRange viewportOfViewer = EditorAPI.getViewport(viewer);

    if (viewportOfViewer == null || document == null) +1 +1
        return; +1

    int lines = document.getNumberOfLines();
    int rangeTop = 0;
    int rangeBottom = 0;
    int selectionTop = 0;
    int selectionBottom = 0;

    if (selection != null) { +1
        try {
            selectionTop = document.getLineOfOffset(selection.getOffset());
            selectionBottom = document.getLineOfOffset(selection
                .getOffset() + selection.getLength());
        } catch (BadLocationException e) { +1
            // should never be reached
            LOG.error("Invalid line selection: offset: "
                + selection.getOffset() + ", length: "
                + selection.getLength());

            selection = null;
        }
    }

    if (range != null) { +1
        if (range.getStartLine() == -1) { +1
            range = null;
        } else {
            rangeTop = Math.min(lines - 1, range.getStartLine());
            rangeBottom = Math.min(lines - 1,
                rangeTop + range.getNumberOfLines());
        }
    }

    if (range == null && selection == null) +1 +1
        return; +1

    // top line of the new viewport
    int topPosition;
    int localLines = viewportOfViewer.getNumberOfLines();
    int remoteLines = rangeBottom - rangeTop;
    int sizeDiff = remoteLines - localLines;

    // initializations finished

    if (range == null || selection == null) { +1 +1
        topPosition = (rangeTop + rangeBottom + selectionTop + selectionBottom) / 2;
        viewer.setTopIndex(topPosition);
        return; +1
    }

    /*
     * usually the viewport of the follower and the viewport of the followed
     * user will have the same center (this calculation). Exceptions may be
     * made below.
     */
    int center = (rangeTop + rangeBottom) / 2;
    topPosition = center - localLines / 2;

    if (sizeDiff <= 0) { +1
        // no further examination necessary when the local viewport is the
        // larger one
        viewer.setTopIndex(Math.max(0, Math.min(topPosition, lines)));
        return; +1
    }

    boolean selectionTopInvisible = (selectionTop < rangeTop + sizeDiff / 2);
    boolean selectionBottomInvisible = (selectionBottom > rangeBottom
        - sizeDiff / 2 - 1);

    if (rangeTop == 0 +1
        && !(selectionTop <= rangeBottom && selectionTop > rangeBottom +1 +1
            - sizeDiff)) {
        // scrolled to the top and no selection at the bottom of range
        topPosition = 0;

    } else if (rangeBottom == lines - 1 +1
        && !(selectionBottom >= rangeTop && selectionBottom < rangeTop +1 +1
            + sizeDiff)) {
        // scrolled to the bottom and no selection at the top of range
        topPosition = lines - localLines;

    } else if (selectionTopInvisible && selectionBottom >= rangeTop) { +1 +1
        // making selection at top of range visible
        topPosition = Math.max(rangeTop, selectionTop);

    } else if (selectionBottomInvisible && selectionTop <= rangeBottom) { +1 +1
        // making selection at bottom of range visible
        topPosition = Math.min(rangeBottom, selectionBottom) - localLines
            + 1;
    }

    viewer.setTopIndex(Math.max(0, Math.min(topPosition, lines)));
}

编辑:我已将复杂性降低到 11。我应该如何进一步降低它？

private int setTopPositionUtil(int sizeDiff, int rangeTop, int rangeBottom, int selectionTop, int selectionBottom) {
boolean selectionTopInvisible = (selectionTop < rangeTop + sizeDiff / 2);
boolean selectionBottomInvisible = (selectionBottom > rangeBottom - sizeDiff / 2 - 1);
if (rangeTop == 0 && !(selectionTop <= rangeBottom && selectionTop > rangeBottom - sizeDiff)) { // +1 +1 +1
    // scrolled to the top and no selection at the bottom of range
    topPosition = 0;
} else if (rangeBottom == lines - 1 && !(selectionBottom >= rangeTop && selectionBottom < rangeTop + sizeDiff)) { // +1 +1 +1
    // scrolled to the bottom and no selection at the top of range
    topPosition = lines - localLines;
} else if (selectionTopInvisible && selectionBottom >= rangeTop) { // +1 +1
    // making selection at top of range visible
    topPosition = Math.max(rangeTop, selectionTop);
} else if (selectionBottomInvisible && selectionTop <= rangeBottom) { // +1 +1
    // making selection at bottom of range visible
    topPosition = Math.min(rangeBottom, selectionBottom) - localLines + 1;
}
return topPosition;
}

private int setTopPosition(int localLines, int rangeTop, int rangeBottom,int selectionTop, int selectionBottom) {

// top line of the new viewport
int topPosition;
int remoteLines = rangeBottom - rangeTop;
int sizeDiff = remoteLines - localLines;

// initializations finished

/*
 * usually the viewport of the follower and the viewport of the followed
 * user will have the same center (this calculation). Exceptions may be
 * made below.
 */
int center = (rangeTop + rangeBottom) / 2;
topPosition = center - localLines / 2;

if (sizeDiff > 0) { // +1
    setTopPositionUtil(sizeDiff, rangeTop, rangeBottom, selectionTop, selec);
}

return Math.max(0, Math.min(topPosition, lines)); // +1
}

private void adjustViewport(IEditorPart editorPart, LineRange range,TextSelection selection) {
ITextViewer viewer = EditorAPI.getViewer(editorPart);
if (viewer != null) { // +1

    IDocument document = viewer.getDocument();
    LineRange viewportOfViewer = EditorAPI.getViewport(viewer);

    if (viewportOfViewer != null && document != null) { // +1 +1

        int lines = document.getNumberOfLines();
        int rangeTop = 0;
        int rangeBottom = 0;
        int selectionTop = 0;
        int selectionBottom = 0;

        if (selection != null) { // +1
            try {
                selectionTop = document.getLineOfOffset(selection.getOffset());
                selectionBottom = document.getLineOfOffset(selection
                    .getOffset() + selection.getLength());
            } catch (BadLocationException e) { // +1
                // should never be reached
                LOG.error("Invalid line selection: offset: " +
                    selection.getOffset() + ", length: " +
                    selection.getLength());

                selection = null;
            }
        }

        if (range != null) { // +1
            if (range.getStartLine() == -1) { // +1
                range = null;
            } else {
                rangeTop = Math.min(lines - 1, range.getStartLine());
                rangeBottom = Math.min(lines - 1,
                    rangeTop + range.getNumberOfLines());
            }
        }

        if (range != null && selection != null) { // +1 +1
            viewer.setTopIndex(setTopPosition(viewportOfViewer.getNumberOfLines(),
                rangeTop, rangeBottom, selectionTop, selectionBottom));
        } else {
            viewer.setTopIndex((rangeTop + rangeBottom + selectionTop + selectionBottom) / 2);
        }
    }
}
}

注意:代码现在由 3 个方法组成。第二个和第三个的复杂度低于 10，但是 setTopPositionUtil 的复杂度仍然是 11。有什么帮助吗？
抱歉缩进。

Answer 1

您的问题非常广泛，但为了让您继续下去，您应该执行以下操作:

1. 为待测代码创建单元测试。
2. 您使用一些覆盖率工具来使测试达到 100% 的覆盖率。

换句话说:您编写了如此多的测试用例，以至于您完全确信您对生产代码中的每个和任何方面都有测试。

然后你开始重构。然后你继续运行你的测试套件；以确保您不会破坏任何东西。并且您继续运行复杂性度量工具；确保您朝着正确的方向前进。

当问题是:我该如何重构时，那就转向Refactoring这样的经典作者:Fowler 或 Clean Code作者:马丁。

总的来说，我认为您也有点关注错误的主题:您的主要目标应该是创建可读代码很容易遵循。我的建议是将这个巨大的方法进一步分割成一些较小的方法。