On Sat, Feb 24, 2018 at 09:38:07AM +0100, Peter Zijlstra wrote: > On Sat, Feb 24, 2018 at 02:30:05PM +0800, Boqun Feng wrote: > > On Sat, Feb 24, 2018 at 01:32:50PM +0800, Boqun Feng wrote: > > > > /* > > > * DEP_*_BIT in lock_list::dep > > > * > > > * For dependency @prev -> @next: > > > * > > > * RR: both @prev and @next are recursive read locks, i.e. ->read == 2. > > > * RN: @prev is recursive and @next is non-recursive. > > > * NR: @prev is a not recursive and @next is recursive. > > > * NN: both @prev and @next are non-recursive. > > > * > > > * Note that we define the value of DEP_*_BITs so that: > > > * bit0 is prev->read != 2 > > > * bit1 is next->read != 2 > > > */ > > > #define DEP_RR_BIT 0 > > > #define DEP_RN_BIT 1 > > > #define DEP_NR_BIT 2 > > > #define DEP_NN_BIT 3 > > > > > > #define DEP_RR_MASK (1U << (DEP_RR_BIT)) > > > #define DEP_RN_MASK (1U << (DEP_RN_BIT)) > > > #define DEP_NR_MASK (1U << (DEP_NR_BIT)) > > > #define DEP_NN_MASK (1U << (DEP_NN_BIT)) > > > > > > static inline unsigned int > > > __calc_dep_bit(struct held_lock *prev, struct held_lock *next) > > > { > > > return (prev->read != 2) + ((next->read != 2) << 1) > > > } > > > > > > static inline u8 calc_dep(struct held_lock *prev, struct held_lock *next) > > > { > > > return 1U << __calc_dep_bit(prev, next); > > > } > > > > > > static inline bool only_rx(u8 dep) > > > { > > > return !(dep & (DEP_NR_MASK | DEP_NN_MASK)); > > > } > > > > > > static inline bool only_xr(u8 dep) > > > { > > > return !(dep & (DEP_NR_MASK | DEP_NN_MASK)); > > > } > > > > > > > > > if (have_xr && is_rx(entry->dep)) > > > > > continue; > > > > > > > > > > entry->have_xr = is_xr(entry->dep); > > > > > > > > > Hmm.. I think this part also needs some tweak: > > > > /* if -> prev is *R, and we only have R* for prev -> this, * skip*/ > > if (have_xr && only_rx(entry->dep)) > > continue; > > > > /* > > * we pick a *R for prev -> this only if: > > * prev -> this dependencies are all *R > > * or > > * -> prev is *R, and we don't have NN for prev -> this > > */ > > entry->have_xr = only_xr(entry->dep) || (have_xr && !is_nn(entry->dep)); > > > > otherwise, we will wrongly set entry->have_xr to false if have_xr is > > true and we have RN for prev -> this. > > OK, so its saturday morning and such, but what? Why should we set > have_xr true when we have RN? Note that if we only had RN we'd already > have bailed on the continue due to only_rx(). > But what if we have RN and NR? only_rx() will return false, but due to have_xr is true, we can not pick RN, so entry->have_xr should be set to true (due to we have to pick NR), however only_xr() is false becuase we have RN, so if we set entry->have_xr to only_xr(), we set it as false. This is for case like: TASK1: read_lock(A); read_lock(B); TASK2: write_lock(B); read_lock(C); TASK3: read_lock(B); write_lock(C); TASK4: read_lock(C); write_lock(A); , which is not a deadlock. Am I missing something sublte? Regards, Boqun > So can you elaborate a bit?