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我有一个带有函数的基类virtual std::ios::openmode defaultMode() const。
virtual std::ios::openmode defaultMode() const
如果我想要:这个函数的语法是什么(defaultMode() | mode) == mode,模式是另一个std::ios::openmode?如何返回std::ios::mode不会修改其他模式的“空”?
(defaultMode() | mode) == mode
std::ios::openmode
std::ios::mode