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  1. PO
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    <p>OK, so you can't do it without a mutex at all, but you can make that mutex fast.</p> <p>First declare a class to hold the mutex, and a ready flag (<code>InitMutex</code> below). When <code>GrabMutex()</code> is called and <code>ready</code> is <code>false</code>, the actual mutex is grabbed. When <code>ReleaseMutex()</code> is called it does the right thing based on the flag sent from <code>GrabMutex()</code>. After the first time through, ready goes true as the static object is now initialized, so the mutex doesn't need to be grabbed.</p> <p>Now, declare a class which calls <code>GrabMutex()</code> in the constructor and <code>ReleaseMutex(flag)</code> in the destructor, saving the flag locally (<code>InitMutexHolder</code> below).</p> <p>Now, instantiate that class inside your regular <code>getSingleton</code> function. This will ensure that the singleton initialization is mutexed the first time through, and if multiple threads contend they will line up on the mutex. But once the singleton is initialized, <code>ready</code> goes true, and access will be fast. The destructor is called magically after <code>return theSingleton</code> is executed, which releases the mutex (or does nothing if the mutex was not taken).</p> <p>But, the main body of code in your <code>theSingleton()</code> function is unchanged, we have only added a control object per singleton, and a stack object in the call which manages the thread safety.</p> <p>Note about write barriers: since the code is safe whenever <code>ready</code> is false, and <code>ready</code> can't go true until the object is initialized, the code is overall safe assuming that writes are instantly visible. But, there may be a delay in <code>ready=true</code> being visible, since there is no write barrier after setting ready true. However, during that delay safety is still maintained, since <code>ready=false</code> is the conservative, safe case.</p> <pre><code>class InitMutex { public: InitMutex() : ready(false) { } bool GrabMutex() { if (!ready) { mutex.Grab(); return true; } else { return false; } } void ReleaseMutex(bool flagFromGrabMutex) { if (flagFromGrabMutex) { mutex.Release(); ready = true; } } Mutex mutex; bool ready; }; class InitMutexHolder { public: InitMutexHolder(InitMutex &amp; m) : initMutex(m) { inMutex = initMutex.GrabMutex(); } ~InitMutexHolder() { initMutex.ReleaseMutex(inMutex); } private: bool inMutex; InitMutex &amp; initMutex; }; static InitMutex singletonMutex; static Singleton &amp; getSingleton() { InitMutexHolder mutexHolder(singletonMutex); { static Singleton theSingleton; return theSingleton; } } </code></pre>
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    1. POMake Meyers' Singleton thread safe and fast with lazy evaluation
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    1. USEddie Edwards
    1. USEddie Edwards
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    1. COI think I forget to add into the topic words "lazy evaluations", sorry.
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    2. COI clearly understand that Singleton is initialized only once, and that initialize it before main will be thread safe, but it won't be lazy evaluation.
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    3. COI know that variant that you wrote now. It is stated in my question in the PS that I'm asking about realization without new and pointers. I guessed before asked my question that the answer is "no, it's not possible", and your post seems to answers the same )
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