Question

Class with two overloaded operator() functions are called from separate threads. See //comments in code below.

Does the optimizer know not to move entryadd = mSpread * ENTRY_MULTIPLIER above the lock()?

struct Algo1
{   
    boost::detail::spinlock mSpreadLock;

    Algo1() : mSpreadLock() {}

    //called from thread 1 
    inline void operator()(const indata &signal) 
    {
        if ( signal.action() == SEND )
        {
            double entryadd;
            mSpreadLock.lock();
            entryadd = mSpread * ENTRY_MULTIPLIER; //isnt it possible for compiler to optimize this before the lock? 
            mSpreadLock.unlock();
            FunctionCall(entryadd);
        }
    }

    //called from thread2
    inline void operator()(const indata2 &bospread) 
    {
        boost::detail::spinlock::scoped_lock mylock(mSpreadLock);
        mSpread = bospread.spread();
    }
}

What about this?

{
    mSpreadLock.lock();
    double entryadd = mSpread * ENTRY_MULTIPLIER; 
    mSpreadLock.unlock();
{

Would the definition of entryadd me moved to top of function?

Unless im missing something.. seems that lock and unlock within a code block will not work. must use scoped_lock. boost::detail::spinlock::scoped_lock mylock(mSpreadLock) , which will lock for the duration of the function call.

Of course I can just hack it like this: (but is less efficient)

inline void operator()(const indata &signal) 
{
    if ( signal.action() == SEND )
    {
        double entryadd;
        {
            boost::detail::spinlock::scoped_lock mylock(mSpreadLock);
            entryadd = mSpread * ENTRY_MULTIPLIER; 
        }
        FunctionCall(entryadd);
    }
}

Answer 1

Locking operations will eventually use compiler built-in functions which perform some type of atomic operation. The compiler knows that those operations must not be reodered and will not optimize “past” them. It’s all fine.