性能表现不佳的处理重复条件重新分配

class BrakemanTest def f x = foo001(x) if bar? x = foo002(x) if bar? x = foo003(x) if bar? x = foo004(x) if bar? x = foo005(x) if bar? x = foo006(x) if bar? x = foo007(x) if bar? x = foo008(x) if bar? x = foo009(x) if bar? x = foo010(x) if bar? x = foo011(x) if bar? x = foo012(x) if bar? x = foo013(x) if bar? x = foo014(x) if bar? x = foo015(x) if bar? x = foo016(x) if bar? x = foo017(x) if bar? x = foo018(x) if bar? x = foo019(x) if bar? x = foo020(x) if bar? x = foo021(x) if bar? x = foo022(x) if bar? x = foo023(x) if bar? x = foo024(x) if bar? x = foo025(x) if bar? x = foo026(x) if bar? x = foo027(x) if bar? x = foo028(x) if bar? x = foo029(x) if bar? x = foo030(x) if bar? x end end