mutex

问题 This question already has answers here : Why is there no need to mark the move constructor of a type with a deleted copy constructor as deleted? (1 answer) Move constructor for std::mutex (2 answers) Closed 6 months ago . Could somebody tell the reasons why std::mutex is neither copyable nor movable? Somebody told me that it has some relationship to avoid resource waste. Why the copy constructor of std::mutex should be marked as deleted? If not, is there any potential problem? Its copy

问题 This question already has answers here : Why is there no need to mark the move constructor of a type with a deleted copy constructor as deleted? (1 answer) Move constructor for std::mutex (2 answers) Closed 6 months ago . Could somebody tell the reasons why std::mutex is neither copyable nor movable? Somebody told me that it has some relationship to avoid resource waste. Why the copy constructor of std::mutex should be marked as deleted? If not, is there any potential problem? Its copy

来源： https://stackoverflow.com/questions/16693992/why-is-sizeofstdmutex-40-gcc-clang-icc

来源： https://stackoverflow.com/questions/52003403/mutex-as-member-of-class

来源： https://stackoverflow.com/questions/44190865/stdtimed-mutextry-lock-for-fails-immediately

来源： https://stackoverflow.com/questions/44190865/stdtimed-mutextry-lock-for-fails-immediately

来源： https://stackoverflow.com/questions/44190865/stdtimed-mutextry-lock-for-fails-immediately

来源： https://stackoverflow.com/questions/44190865/stdtimed-mutextry-lock-for-fails-immediately

问题 Let's assume we have a memory area where some thread is writing data to. It then turns its attention elsewhere and allows arbitrary other threads to read the data. However, at some point in time, it wants to reuse that memory area and will write to it again. The writer thread supplies a boolean flag ( valid ), which indicates that the memory is still valid to read from (i.e. he is not reusing it yet). At some point he will set this flag to false and never set it to true again (it just flips

问题 I have the following code structure where I Lock() at point A and need to definitely Unlock() at a point B in the same function. Between point A and B, I have multiple returns based on errors where I would need to Unlock(). Using defer lock.Unlock() at point A solves the problem that in case there are errors the lock would be released. However, if execution successfully reaches point B - how can I cancel that defer and Unlock() anyway? func foo() { ... // point A lock.Lock() defer lock.Unlock