如何降低圈复杂度13（超阈值10）的isPropertyAccessed方法复杂度

• 现有isPropertyAccessed方法圈复杂度检测值为13，高于允许阈值10
• 优化约束：必须保留方法内全部判断条件，所有条件无重复逻辑，不得删减业务判断规则
  待优化原代码如下：

private boolean isPropertyAccessed(AccessLevels readAccessLevel, Relationship relationship) {
    boolean isAccessed1 = readAccessLevel.isAccessed1() && relationship.getAccessed1Relationship() == ACCESSED1;
    boolean isAccessed2 = readAccessLevel.isAccessed2() && relationship.getAccessed2Relationship() == ACCESSED2;
    boolean isAccessed3 = readAccessLevel.isAccessed3() && relationship.getAccessed3Relationship() == ACCESSED3;
    boolean isAccessed4 = readAccessLevel.isAccessed4() && relationship.getAccessed4Relationship() == ACCESSED4;
    boolean isAccessed5 = readAccessLevel.isAccessed5() && relationship.getAccessed5Relationship() == ACCESSED5;
    boolean isAccessed6 = readAccessLevel.isAccessed6() && relationship.getAccessed6Relationship() == ACCESSED6;
    return readAccessLevel.isAll() || isAccessed1 || isAccessed2 || isAccessed3
        || isAccessed4 || isAccessed5 || isAccessed6;
}

圈复杂度统计规则会将每个逻辑与（&&）、逻辑或（||）、条件分支、循环都计为独立分支节点：原代码包含1个基础执行路径、6个&&判断、6个||判断，合计复杂度为1+6+6=13，和检测结果一致。

优先推荐延迟判断+短路遍历实现，在绝大多数业务场景下（判断逻辑无副作用）可以做到逻辑等价、性能更优，且圈复杂度降到4，远低于阈值要求。

推荐实现（无副作用场景，性能更优） #

private boolean isPropertyAccessed(AccessLevels readAccessLevel, Relationship relationship) {
    // 优先判断全权限标记，命中则直接返回，避免后续无用判断
    if (readAccessLevel.isAll()) {
        return true;
    }
    // 严格按原代码顺序封装判断逻辑为延迟执行单元，保证判断优先级与原逻辑一致
    List<Supplier<Boolean>> accessChecks = Arrays.asList(
        () -> readAccessLevel.isAccessed1() && relationship.getAccessed1Relationship() == ACCESSED1,
        () -> readAccessLevel.isAccessed2() && relationship.getAccessed2Relationship() == ACCESSED2,
        () -> readAccessLevel.isAccessed3() && relationship.getAccessed3Relationship() == ACCESSED3,
        () -> readAccessLevel.isAccessed4() && relationship.getAccessed4Relationship() == ACCESSED4,
        () -> readAccessLevel.isAccessed5() && relationship.getAccessed5Relationship() == ACCESSED5,
        () -> readAccessLevel.isAccessed6() && relationship.getAccessed6Relationship() == ACCESSED6
    );
    for (Supplier<Boolean> check : accessChecks) {
        if (check.get()) {
            return true;
        }
    }
    return false;
}

实现说明 #

• 所有业务判断规则100%保留，判断优先级、短路逻辑和业务语义与原代码对齐
• 性能优于原代码：全权限场景下直接返回，不需要执行6组访问判断；非全权限场景下按顺序判断，命中即返回，不会执行后续多余判断
• 复杂度达标：方法内仅包含2个条件分支、1个循环分支，加基础路径总圈复杂度为4，符合阈值要求

严格等价实现（保留原代码执行顺序） #

如果原代码中访问判断的方法调用存在副作用（如计数、日志、异常抛出等特殊逻辑），需要完全保留原代码的执行顺序和调用次数，可以使用以下实现，圈复杂度为10，刚好达到阈值要求：

private boolean isPropertyAccessed(AccessLevels readAccessLevel, Relationship relationship) {
    // 按原代码顺序先执行所有访问判断，保留原有调用逻辑
    boolean isAccessed1 = readAccessLevel.isAccessed1() && relationship.getAccessed1Relationship() == ACCESSED1;
    boolean isAccessed2 = readAccessLevel.isAccessed2() && relationship.getAccessed2Relationship() == ACCESSED2;
    boolean isAccessed3 = readAccessLevel.isAccessed3() && relationship.getAccessed3Relationship() == ACCESSED3;
    boolean isAccessed4 = readAccessLevel.isAccessed4() && relationship.getAccessed4Relationship() == ACCESSED4;
    boolean isAccessed5 = readAccessLevel.isAccessed5() && relationship.getAccessed5Relationship() == ACCESSED5;
    boolean isAccessed6 = readAccessLevel.isAccessed6() && relationship.getAccessed6Relationship() == ACCESSED6;

    if (readAccessLevel.isAll()) {
        return true;
    }
    // 用循环遍历替代连续||判断，减少分支计数
    for (boolean accessedResult : new boolean[]{isAccessed1, isAccessed2, isAccessed3, isAccessed4, isAccessed5, isAccessed6}) {
        if (accessedResult) {
            return true;
        }
    }
    return false;
}

不推荐将6个判断拆分为独立私有方法的实现方式，虽然也能降低当前方法的圈复杂度，但会增加代码跳转成本，可读性没有上述两种方案好。

内容的提问来源于stack exchange，提问作者Valeriy K.