How can I device an algorithem for deadlock detection through an Injecting DLL containing the hooked Functions for locking and Unlocking.
Actually I want to device and algorithem for deadlock detection. Kindly 开发者_StackOverflow社区if someone could me in this regard.
Conceptually deadlock detection is simple in principle, but hard to implement.
From a high-level point of view, what you want is to record the locks held by each thread and see if acquiring the lock would result in a dead-lock. You can visualize this with a dependency graph, a cycle meaning a deadlock.
However, there are other operations that you can use for synchronizations: spin-locking for example. Those will screw up any attempt on detection, so be aware of the restrictions.
So let's have a simulation first: Imagine 3 threads (T1, T2, T3) and 3 mutexes (M1, M2, M3)
- T1 grab M1
- T2 grab M2, wait for M1
- T3 grab M3, wait for M2
If T1
waits for M3
you're screwed (you have a cycle), thus before trying to grab, you need to check for this condition.
You can modelize this using:
- a table, which lists the threads holding a given mutex
- a graph, representing the dependencies between threads
If we modelize the situation when T1
tries to grab M3
we have:
Table
M1 -> T1,
M2 -> T2,
M3 -> T3,
Graph
{T1, T2, T3} x {T2 -> T1, T3 -> T2}
When T1
tries to grab M3
:
- It looks up the table and list the threads holding it, here
T3
. - It checks if adding the edge
T1 -> T3
in the graph forms a cycle.
See this CP article - you're not original I'm afraid. See also this microsoft.public.win32.programmer.kernel article where a Microsoft employee explains the WIndows built-in options.
A simple algorithm is making a Wait-For graph for the resource allocation...
for this u maintain a structure like
struct process
{
int process_id;
int curr_res;
int want_in_future;
};
the representation will be....
(R1)----->[P1]------>(R2)
meaning is ... R1 has been allocated to P1 and now this process will be desiring to have R2 in future
but this choice may lead u to DEADLOCK like
for p1 (R1)----->[P1]------>(R2)
for p2 (R2)----->[P2]------>(R1)
a circular way is there so deadlock will be there
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