C++ Concurrent profiling using Helgrind - a tool of Valgrind

 On 15th August - my contribution to the learning community...

Concurrency profiling in C++ is essential for optimizing the performance of multi-threaded applications by identifying and addressing bottlenecks, inefficiencies, and issues like race conditions and deadlocks.

I am using the Helgrind tool of Valgrind in eclipse to do the experimentation on C++ data race condition in a multithreaded application.

A data race occurs in a multithreaded application when two or more threads access shared data concurrently, and at least one of these accesses is a write operation without proper synchronization (e.g., without locks). This can lead to unpredictable behavior, crashes, or incorrect program output.

Here is some information about Helgrind.

Helgrind:

- Detects data races, potential deadlocks, and lock-order violations.

- Useful for debugging multi-threaded applications where data consistency is crucial.

Data Race Detection: 

If Helgrind detects that two threads are accessing the same memory location concurrently without proper synchronization, and at least one of these accesses is a write, it flags this as a data race.

Please have a look at my video - it's all explained here.

Concurrency profiling in C++ is crucial for developing high-performance multi-threaded applications.

Valgrind is a versatile tool for detecting a wide range of memory-related issues in C++ applications. Tools like Memcheck, Helgrind, DRD, and Massif provide comprehensive coverage of memory leaks, invalid accesses, uninitialized memory usage, threading issues, and memory management inefficiencies.

Using Valgrind in the development cycle can significantly improve the stability and performance of your application by identifying and allowing you to fix these memory-related issues.

Another important task that Helgrind does is to check whether there is any deadlock in a multithreaded C++ application - like Cyclic dependency.

Cyclic dependency deadlock occurs when two or more locks are acquired in a different order in two task executions, potentially leading to a deadlock when the program's tasks execute in parallel.

A Lock order violation problem indicates the following timeline:

Task 1

Acquire lock A.

Acquire lock B.

Release lock B.

Release lock A.

Task 2

Acquire lock B.

Acquire lock A.

Release lock A.

Release lock B.

If these time lines are interleaved when the two tasks execute in parallel, a Deadlock occurs:

Task 1: Acquire lock A.

Task 2: Acquire lock B.

Task 1: Try to acquire lock B; wait until task 2 releases it.

Task 2: Try to acquire lock A; wait until task 1 releases it.

For example consider the following piece of code

//============================================================================

// Name : CyclicDependencyDeadLock.cpp

// Author : Som

// Version :

// Copyright : som-itsolutions

// Description : Hello World in C++, Ansi-style

//============================================================================

#include <iostream>

#include <thread>

#include <mutex>

using namespace std;

std::mutex lock1, lock2;

void threadA() {

int count = 0;

std::lock_guard<std::mutex> guard1(lock1);

for (int i = 0; i<100000; i++){

count++;

}

std::lock_guard<std::mutex> guard2(lock2);

//cout<<"Thread id " <<this_thread::get_id()<<" this is okay now because of correct lock order " <<count<<endl;

cout<<"Thread id " <<this_thread::get_id()<< " this will never be printed..."<<count<<endl;

}

void threadB() {

std::lock_guard<std::mutex> guard2(lock2);

int count = 0;

for (int i = 0; i<100000; i++){

count++;

}

std::lock_guard<std::mutex> guard1(lock1);

//cout<<"Thread id " <<this_thread::get_id()<<" this is okay now because of correct lock order " <<count<<endl;

cout<<"Thread id " <<this_thread::get_id()<< " this will never be printed..."<<count<<endl;

}

int main() {

std::thread t1(threadA);

std::thread t2(threadB);

t1.join();

t2.join();

return 0;

}

And if we profile the above piece of code using Helgrind, it will show

Thread #3: lock order "0x10E160 before 0x10E1A0" violated...

Please have a look at the following video...

To avoid the lock order violation we must go for consistent global order for all the threads.

That's all for today...

I hope this exploration will help the inquisitive minds of software engineers.

Jai Hind.... Jai Bharat...