> I find a line in tbf_change() in sch_tbf.c: ptab = xchg(&q->P_tab, ptab); > > Can I consider that the pointers are exchanged between q->P_tab and ptab? as following: > > prev=q->R_tab; > q->R_tab = rtab; > rtab = prev; > > Is my understanding righ? Yes almost. old = xchg(ptv,v) does old=*ptr; *ptr=v and does it atomicaly (!) thus no locking is needed. devik
