Subj : Re: recursive mutexes
To   : comp.programming.threads
From : Peter Dimov
Date : Sun May 15 2005 07:11 pm

Uenal Mutlu wrote:
> "Peter Dimov" wrote
>> Uenal Mutlu wrote:
>>> "David Schwartz" wrote
>>>>     Yes, it can and must know that it already has the lock. A
>>>> function that operates on X should be defined as being called with
>>>> X locked, or with X unlocked. It could even take a boolean that
>>>> indicates whether the caller has locked the object or not, though
>>>> this is rarely the right thing to do.
>>>
>>> You are unnecessarily complicating things and make the program
>>> slower by doing these testings.
>>
>> No, he's in fact making the program faster.
>
> Are you sure? Let's see:
>
> void X::f()
> {
>   Lock();
>   for (int i = 0; i < 1000; i++)
>     f();

Stack overflow or deadlock here, you probably wanted to call g().

>   Unlock();
> }
>
> void X::g()
> {
>  if (!IsLocked())
>    return;
>  //...do..something...
> }
>
> Are you saying the above version of g() is faster than this one:  ?
> void X::g()
> {
>  //...do..something...
> }

Of course not. Why would you want to test anything if you could just use the 
above? We're talking about recursive mutexes, remember?

void X::f()
{
    Lock();

    for (int i = 0; i < 1000; i++)
    {
        g();
    }

    Unlock();
}

void X::g()
{
    Lock();

    //...do..something...

    Unlock();
}

compared to

void X::f()
{
    Lock();

    for (int i = 0; i < 1000; i++)
    {
        g( false );
    }

    Unlock();
}

void X::g( bool lock )
{
    if( lock ) Lock();

    //...do..something...

    if( lock ) Unlock();
}

Let's overlook the fact that the correct version is of course:

void X::f()
{
    Lock();


    for (int i = 0; i < 1000; i++)
    {
        g_unlocked();
    }

    Unlock();
}


void X::g_unlocked()
{
    //...do..something...
}

void X::g()
{
    Lock();
    g_unlocked();
    Unlock();
}

.