C++ Core Guidelines: Concurrency and lock-free Programming


Today, I finish the rules to concurrency and continue directly with lock-free programming. Yes, you have read it correctly: lock-free programming.

 fear cat

Before I write about lock-free programming in particular, here are the three last rules to concurrency.

I make it short because these rules are quite obvious.

CP.43: Minimize time spent in a critical section

The less time your lock a mutex, the more time other threads can run. Have a look at the notification of a condition variable. If you want to see the entire program, read my previous post C++ Core Guidelines: Be Aware of the Traps of Condition Variables.

void setDataReady(){
    std::lock_guard<std::mutex> lck(mutex_);
    dataReady = true;                                // (1) 
    std::cout << "Data prepared" << std::endl; 


The Mutex mutex_ is locked at the beginning of the function and unlocked at the end of the function. This is not necessary. Only the expression dataReady = true (1) has to be protected.

First, std::cout is thread-safe. The C++11 standard guarantees that each character is written in an atomic step and in the right sequence. Second, the notification condVar.notify_one() is thread-safe.

Here is the improved version of the function setDataReady:

void setDataReady(){
    {  // Don't remove because of the lifetime of the mutex (1)
        std::lock_guard<std::mutex> lck(mutex_);
        dataReady = true;
    }                                                       (2)
    std::cout << "Data prepared" << std::endl;


When I teach this rule in my classes to concurrency, often a question arises:  Should you document this artificial scope for limiting the lifetime of the std::lock_guard and, therefore, the lifetime of the std::mutex (line (1) and (2))? Most of my students including me add a comment to this artificial scope. If not, the maintainer of your code may not be aware that the curly braces in line (1) and line (2) are necessary for control the lifetime of the mutex. In the end, your maintainer will remove the artificial scope because they seem to be superfluous and your code uses the first variant.


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CP.44: Remember to name your lock_guards and unique_locks

I was a little astonished to read this rule. Here is the example from the guidelines:

lock_guard<mutex> {m2};
lock(m1, m2);


The unique_lock and lock_guard are just temporaries that are created and immediately destroyed. The std::lock_guard or std::unique_lock locks its mutex and its constructor and unlocks it in its destructor. This pattern is called RAII. Read the details here: Garbage Collection: No Thanks.

My small example shows only the conceptional behaviour of a std::lock_guard. Its big brotherstd::unique_lock supports more operations.


// myGuard.cpp

#include <mutex>
#include <iostream>

template <typename T>
class MyGuard{
  T& myMutex;
    MyGuard(T& m):myMutex(m){
	  std::cout << "lock" << std::endl;
      std::cout << "unlock" << std::endl;

int main(){
  std::cout << std::endl;
  std::mutex m;
  MyGuard<std::mutex> {m};                        // (1)
  std::cout << "CRITICAL SECTION" << std::endl;   // (2)
  std::cout << std::endl;

}                                                 // (3)


MyGuard calls lock and unlock in its constructor and in its destructor. Because of the temporary, the call to the constructor and destructor happens in line (1). In particular, this means that the call of the destructor happens at the line (1) and not, as usual, in line (3). As consequence, the critical section in line (2) is executed without synchronisation.

This execution of the program shows that the output of "unlock" happens before the output of "CRITICAL SECTION".


CP.50: Define a mutex together with the data it guards. Use synchronized_value<T> where possible

The central idea is that you put your mutex into the data you want to protect. With already standardised C++ it looks like this:

struct Record {
    std::mutex m;   // take this mutex before accessing other members
    // ...


With an upcoming standard, it may look like this because synchronized_value<T> is not part of the current C++ standard but it may become part of an upcoming standard.

class MyClass {
    struct DataRecord {
       // ...
    synchronized_value<DataRecord> data; // Protect the data with a mutex


According to the proposal N4033 from Anthony Williams: "The basic idea is that synchronized_value<T> stores a value of type T and a mutex. It then exposes a pointer interface, such that dereferencing the pointer yields a special wrapper type that holds a lock on the mutex, and that can be implicitly converted to T for reading, and which forwards any values assigned to the assignment operator of the underlying T for writing."

This means that the operation on s in the following code snippet is thread-safe.

synchronized_value<std::string> s;

std::string readValue()
    return *s;

void setValue(std::string const& newVal)
void appendToValue(std::string const& extra)

Now as announced, to something completely different: lock-free programming.

Lock-free programming

First, let me state the most important meta-rule to lock-free programming.

Don't program lock-free

Sure, you don't believe me but based on my experience giving many concurrency classes and workshops this is my first rule. Honestly, I agree with many of the most appreciated C++ experts worldwide. Here are the key statements and cites from their talks:

  • Herb Sutter: Lock-free programming is like playing with knives.
  • Anthony Williams: "Lock-free programming is about how to shoot yourself in the foot."
  • Tony Van Eerd: "Lock-free coding is the last thing you want to do."
  • Fedor Pikus: "Writing correct lock-free programs is even harder."
  • Harald Böhm: "The rules are not obvious."

Here is a picture to the statements and cites:


You still don't believe me? With C++11 the memory order std::memory_order_consume was defined. Seven years later the official wording is: "The specification of release-consume ordering is being revised, and the use of memory_order_consume is temporarily discouraged." (memory_order)

In case you know what you do, think about the ABA problem in the guidelines CP.100.

CP.100: Don’t use lock-free programming unless you absolutely have to

The following code snippet from the C++ core guidelines has a bug.

extern atomic<Link*> head;        // the shared head of a linked list

Link* nh = new Link(data, nullptr);    // make a link ready for insertion
Link* h = head.load();                 // read the shared head of the list

do {
    if (h->data <= data) break;        // if so, insert elsewhere
    nh->next = h;                      // next element is the previous head
} while (!head.compare_exchange_weak(h, nh));    // write nh to head or to h


Find the bug and write me an e-mail. I will mention the best problem analysis in my next post and if you like your name.

What's next?

Of course, I solve the riddle of the ABA problem in the next post. Afterwards, my story with lock-free programming continues. 




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Tags: lock-free


0 #1 Lubos Matejicek 2018-11-28 11:54
In CP.43 is probably the problem... look at this:

void setDataReady(){
{ // Don't remove because of the lifetime of the mutex (1)
std::lock_guard lck(mutex_);
dataReady = true;
} (2)
0 #2 mikethomson 2021-01-21 13:44
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0 #3 Ben Stokes 2021-01-21 14:33
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-2 #4 hitachi 1.5 ton 2021-06-12 06:16
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