Wait until condition is not met - but why mutex?

[TomNow99]

Hello,
I have 2 threads. I would like something like:
main Thread wait when special condition will be met ( for example variable "a" will be 6 )
the second Thread do something and set variable "a"

mainThread:
another operations
waitUntilVariableAIsNot6
another operations

second Thread:
another operations
a = 6;
another operations

I find very good class for that: QWaitCondition
So I would like to write:

mainThread:
another operations
waitCon.wait();
another operations

second Thread:
another operations
a = 6;
waitCon.wakeAll();
another operations

My code is not good. In QWaitCondition we have wait() function, but one of the parameter is "QMutex *lockedMutex". Why I have to use mutex in this example?

[SGaist]

Hi,

Because you want to access a shared resource and thus its access must be protected.

[TomNow99]

@SGaist But I know I use only once variable "a" in second Thread.

EDIT

I agree with you. My example is very uncommon. But why we can't have something like:

mainThread:
another operations
waitCon.wait();
mutex.lock();
another important operations
mutex.unlock();
another operations

second Thread:
another operations
mutex.lock();
a = 6;
mutex.unlock();
waitCon.wakeAll();
another operations

And function wait() can be without parameter "QMutex *lockedMutex".

[SGaist]

In that case it's simply used as a mean to stop the thread execution until the wait condition is awaken.

Depending on what you want to implement, using signals and slot might be a better tool.

[TomNow99]

@SGaist I can't find function like wait() when I want to use signals and slots. Of course in slot I can set variable a.

[SGaist]

Because there are none.

Can you explain what your two threads are doing ? Especially why the first one needs to wait for a to be equal to 6 ?

[TomNow99]

@SGaist I don't have at the moment very good example Sometimes I have code:

fun()
{
fun1();
fun2();
fun3();
}

void timerTimeoutSlot()
{
do Something // for example set very important Variable;
}

I would like to something like:

fun()
{
fun1();
wait when the very important variable will be set
fun2();
fun3();
}

Because when this variable will not set my fun2() can't be execute.

[SGaist]

And fun should be called repeatedly ?

[TomNow99]

@SGaist said in Wait until condition is not met - but why mutex?:

repeatedly

When I look at my code I think I can do something like:

void timerTimeoutSlot()
{
do Something // for example set very important Variable;
fun2();
fun3();
}

But Could you tell me what is going on when signal is emitted? I talk about only 1 thread. For example:

function()
{
operation a;
operation b;
operation c;
operation d;
operation e;
operation f;
}

timerTimeoutSlot()
{
operation m;
operation n;
operation o;
}

timer2TimeoutSlot()
{
operation x;
operation y;
operation z;
}

And when the operation c is execute, signal timeout in timer1 is emitted
So operations will be execute like:
a -> b -> c -> m -> n -> o -> d -> e -> f?

And second question:
When the operation c is execute, signal timeout in timer1 is emitted and when operation m is execute, signal timeout in timer 2 is emitted. So operation will be execute like:
a -> b -> c -> m -> x->y->z->n -> o -> d -> e -> f?

EDIT:
I hear that can be something like:
a -> b -> c -> m -> d->n->e->o->f ( in first question )

[SGaist]

If single threaded, the slots will be called after the functions are done in order.

Looks like you are trying to model some sort of behaviour. What would that be ?

[qwe3]

@SGaist so will be a b c d e f m n o
Or
A b c m d n e o f?

I am a newbie programmer, so my questions are not very smart and I dont have any plans.

[SGaist]

Your answers makes it look like you are trying to model some behaviour that could be represented by a state machine rather that complex threading.

Since you are starting in the programming field, you should rather stay away from the threading paradigm and learn the rest first. Threading is a very complex topic that offers you many way to shoot yourself in the foot from different unrelated angles at different unrelated time with different unrelated side-effects that will want to go after your foot again.

[qwe3]

@SGaist Ok :) I will learn thread after easier thing. But please answer for my question about operations. Thank you

[SGaist]

If timer2 timeout value is greater than timer they will be in order: function -> timerTimeoutSlot -> timer2TimeoutSlot.