mysql deadlock

Deadlock analysis 1, the production of deadlocks There is the following code, which simulates the transfer between two accounts void Transfer (account From,account To,int money) { from.setamount (From.getamount ()-money); To.setamount (To.getamount () +money);} Single-threaded, this code is sure to be okay, but there are problems with multithreading Now lock it, lock the code as follows void Transfer (account From,account To,int money) { syn

(reproduced in Breaksoftware's CSDN blog)Previous articles have explained the causes of deadlocks created in DllMain and waiting for threads to cause. Remember, we analyzed the half-day assembly before we knew the deadlock location in the thread. If the lack of debugging experience for students, may find this location a bit troublesome. So this article will introduce a few examples, they will be the obvious location of the thread deadlock.The code in

Tags: using SP data BS Algorithm database nbsp Method how toThe main causes of deadlocks are:(1) Due to insufficient system resources.(2) The sequence of progress of the process is not appropriate.(3) Improper allocation of resources and so on.The four necessary conditions for creating a deadlock:(1) Mutex condition: A resource can only be used by one process at a time.(2) Request and hold condition: When a process is blocked by a request for resource

One, deadlockDeadlock refers to a situation like this. When multiple threads compete for scarce resources, it is impossible for them to continue because they are waiting for the resources owned by each other's threads.The P2 process has R1 resources, but he is asking for R2 resources, while the P1 process has R2 resources, but he is requesting R1 resources.1.1 Coffman conditions--4 conditions for generating a deadlockIf the following 4 scenarios exist in a system, the chance of a

/* Deadlock, handwritten deadlock Conditions of the deadlock formation: Request to remain The process has maintained at least one resource, but a new resource request has been made, and the resource is already occupied by other processes. At this point, the process is blocked, but the resource you are getting is kept. Loop wait There is a circular wait chain for

) { WaitForSingleObject (G_hevent, INFINITE); return 0; } BOOL apientry DllMain ( hmodule hmodule, DWORD Ul_reason_for_call, lpvoid lpreserved ) { switch (ul_reason_for_call) { Case Dll_process_attach: { G_hevent =:: CreateEvent (NULL, FALSE, FALSE, NULL); G_thread_handle =:: CreateThread (NULL, 0, threadcreateindllmain,null, 0, NULL); } break; Case Dll_process_detach: { :: SetEvent (G_hevent); :: WaitForSingleObject (G_thread_handle, INFINITE);

A deadlock occurs in the program, which causes the w3wp process to restart. This problem is usually difficult to find the cause. I know two instances that may cause this problem. 1. lock or ReaderWriterLock is used in the program, and resource lock contention occurs.Below is a short piece of code: Copy codeThe Code is as follows: // _ rwLock is of the ReaderWriterLock type. _ RwLock. AcquireWriterLock (100 ); DoSomething (); _ RwLock. ReleaseWriterLo

Find out the source of deadlock and blocking for SQL Server--Find out the source of SQL Server deadlock and blocking Copy Code code as follows: Use master Go declare @spid int, @bl int DECLARE S_cur CURSOR for Select 0, blocked From (SELECT * from sysprocesses where blocked>0) a Where NOT EXISTS (SELECT * from sysprocesses where blocked>0) b where A.blocked=spid) Union selec

Deadlock Overview Typically, the simplest and most common deadlocks occur at different table levels,Session 1 The first step to modify a table, the second step to modify the B table,Session 2 The first step to modify the B table, the second step to modify a table,When the session 1 and Session 2 push sequence occurs when the deadlock occurs, this method of ending is relatively simple, in the same orde

Like debugging the lock in userspace, it mainly involves: 1. Locate the lock deadlock and print out the call stack. 2. Find out who is using this mutex. For example, in kernel config, enable is associated with lock/mutex, such as config_debug_mutex and config_detect_hung_tasks. In this way, when a deadlock occurs, wait for a period of time (120 seconds by default). The kernel detects the

---------------------------------- If you write database products in the future, instead of writing SQL statements, there will be more locks involved. There are five philosophers in the story. These philosophers are very poor and can only afford Five chopsticks. They sat in a circle and put a pair of chopsticks in the middle. Philosophers must have chopsticks on both the left and right sides when eating. If one of the people around him is using chopsticks, he will only have to wait. Ass

First, the concept of deadlock Synchronization of multiple threads can cause deadlocks if they are poorly designed. A deadlock is when multiple threads share certain resources while waiting for the other to release resources, causing the program to stagnate. Deadlock can cause the program to stall, so we must be careful to avoid deadlocks when we write multi-t

 Package test.thread;public class Deadlocktest extends thread{Private Object o;public void SetO (Object o) {This.o=o;}public void Run () {Synchronized (this) {SYSTEM.OUT.PRINTLN ("deadlocktest begin");Synchronized (o) {System.out.println ("Deadlocktest_o");}System.out.println ("Deadlocktest end");}}public static void Main (string[] args) {Deadlocktest dlt=new deadlocktest ();TT tt=new TT (DLT);Dlt.seto (TT);Dlt.start ();Tt.start ();}}Class TT extends thread{Private Object o;Public TT (Object o

PackageCom.swift; Public classDeadLockImplementsRunnable {Private BooleanFlag; DeadLock (Booleanflag) { This. Flag =Flag; } Public voidrun () { while(true) { if(flag) {synchronized("Suo1") {System.out.println (Thread.CurrentThread (). GetName ()+ "If....locka"); synchronized("Suo2") {System.out.println (Thread.CurrentThread (). GetName ()+ "IF.....LOCKB"); } } } Else { synchronized("Suo2"

Steps for canceling the deadlock check code and canceling the deadlock session in the EBS backend --- Experience 1. General process for handling request congestion: 1. Execute cc_post_hang. SQL in PL/SQL (to Tao ); [SQL] ----------------------------------------- -- lock -- cc_post_hang. SQL select t. INST_ID, gs. SID, gs. status, gs. MACHINE, gs. ACTION, gs. MODULE, gs. PROGRAM, gs. p3TEXT, gs. EVENT, t. TY

', 4, ' Share ', 5, ' s/row-x ', 6, ' Exclusive ', ' Unknown ') Requestmode, CTime, Block B, V$session.username,machine,module,action, decode (a.type, ' MR ', ' Media Recovery ', ' RT ', ' Redo Thread ', ' UN ', ' User Name ', ' TX ', ' Transaction ', ' TM ', ' DML ', ' UL ', ' PL/SQL User Lock ', ' DX ', ' Distributed Xaction ', ' CF ', ' Control File ', ' is ', ' Instance state ', ' FS ', ' File Set ', ' IR ', ' Instance Recovery ', ' ST ', ' Di SK Space Transaction ', ' TS ', ' Temp Segment

; Select V$session.sid,v$session. Serial#,v$process.spid, RTrim (object_type) Object_type,rtrim (owner) | | ‘.‘ || object_name object_name, decode (lmode, 0, ' None ', 1, ' Null ', 2, ' Row-s ', 3, ' Row-x ', 4 , ' Share ', 5, ' S/row-x ', 6, ' Exclusive ', ' Unknown ') Lockmode, decode (request, 0, ' None ', 1, ' Null ', 2, ' Row-s ', 3, ' Row-x ', 4, ' Share ', 5, ' S/row-x ', 6, ' Exclusive ', ' Unknown ') Requestmode, CTime, Block B, V$session.username,machine,module,action, Decode

,rtrim (owner) | | '. ' | | object_name object_name, decode (lmode, 0, ' None ', 1, ' Null ', 2, ' Row-s ', 3, ' Row-x ', 4, ' Share ', 5, ' S/row-x ', 6, ' excluSive ', ' Unknown ') Lockmode, decode (request, 0, ' None ', 1, ' Null ', 2, ' Row-s ', 3, ' Row-x ', 4, ' Share ', 5, ' S/row-x ', 6, ' Ex Clusive ', ' Unknown ') Requestmode, CTime, Block B, V$session.username,machine,module,action, decode (a.type, ' MR ', ' Media Recovery ', ' RT ', ' Redo Thread ', ' UN ', ' User Name ', ' TX ', ' T

If there are two locks Lock1 (Mutex_gard mode) and lock;The order of the two calls is different, there will be waiting for each other, resulting in deadlock;To avoid deadlocks, we can:1, each thread lock in the same order of invocation;2, the use of std:: Lock ();In a method called by both the main thread and the child threadStd::lock (Mutex1,mutex2,....);When it's locked.In addition: The mutex itself also has the lock and unlock methodC + + multithre

The problem, literally, is that the program has a deadlock, which causes the w3wp process to restart. Usually this problem is difficult to find out the reason. I know two instances that could cause this problem 1. Lock or ReaderWriterLock is used in the program, and the locking resource has been contention Here's a little piece of code: Copy Code code as follows: The type of _rwlock is ReaderWriterLock _rwlock.acquirewriterlock (10