Doc > Qt 4.8 > Mandelbrot Example

[dm_kiselev]

In this example Mandelbrot Example | Qt 4.8, in function RenderThread::run() why we don't protect the image member by a mutex? We write image here and read it in other thread, in function MandelbrotWidget::updatePixmap(const QImage &image, double scaleFactor). So why its not conflict?

[SGaist]

Hi and welcome to devnet,

Because each time you pass in the loop a new image is created so you are not accessing a shared resource.

[dm_kiselev]

@SGaist you mean forever loop or while (pass < NumPasses) loop?

[SGaist]

The forever loop contains the while (pass < NumPasses)

[dm_kiselev]

@SGaist said:

each time you pass in the loop a new image is created

What loop you mean here?

[SGaist]

The forever loop

[dm_kiselev]

Ok, but after emitting the signal we go to the next iteration of the while (pass < NumPasses) and write to the same image. And we read it in another thread at the same time.

[SGaist]

From the documentation of the example: "With queued connections, Qt must store a copy of the arguments that were passed to the signal so that it can pass them to the slot later on."

[dm_kiselev]

Thank you!

[SGaist]

You're welcome !

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