On 2010-08-18 15:53:49 +0200, Victor Bazarov said:
On 8/18/2010 8:37 AM, Philipp Kraus wrote:
I'm to create a multithreaded algorithm and I get this runtime error
message:
incorrect checksum for freed object - object was probably modified
after being freed
[...]
With one thread everything works fine, on more threads I'll get the
error message. I understand this correctly, that the message says
that several threads on another object to access and modify data
there?
The statement after the last 'that' is supposed to be complete.
Let's try to use it as a separate sentence:
"Several threads on another object to access and modify data there"
"Several threads" is the subject.
"Another object" is the object (with "on" as the preposition).
Where is the verb? Several threads WHAT on another object?
Sorry I'm a little bit confused
I would like to understand the message in that way, that I can solve
my problem
If you pass some object by reference to any threads, make sure that
the object *survives* while there is at least one thread using it.
The best approach is the reference counting. Increment the counter
just before a thread begins using the object. When the thread is
done using that object, *the thread* must decrement the reference
counter. When the counter goes to zero, the object should delete
itself.
Here a code excerpt:
std::vector< worker<T,D> > l_worker;
boost::thread_group l_threadgroup;
for(std::size_t i=0; i < p_threads; ++i) {
l_worker.push_back( worker<T,D>(p_iteration, p_stepsize,
m_fulltable, m_derivation, m_optimize, m_static, p_batch) );
l_threadgroup.create_thread( boost::bind(
&worker<T,D>::optimize, l_worker[i] ) );
}
My worker objects get parameters, which are only read in ::optimize. I
don't change the data in the parameter objects (or changes the
iterator the object?). If I copy the reference by the copy-constructor
or std::copy to a new object in my worker / thread object, I wouldn't
get the error message? My thread object throws the error message on
different runs with a different number of messages, but the numer is
less than thread count.