Implementing bluetooth - general questions - part 3 - trace bluetooth inquiry in Qt

[AnneRanch]

I am implementing Qt example "btscanner" .
I am having an issue with consistently detecting working / powered up nearby bluetooth devices. The key word is consistently.

Scanning for nearby bluetooth device is standard function implemented in "blueZ' stack - sometime named "inquiry". Qt implements such scan via
QBluetoothDeviceDiscoveryAgent class and it is physically started by
"using" QBluetoothDeviceDiscoveryAgent WHAT (?) declared as

public Q_SLOTS:
void start();

That is as far as tracing the Qt source code goes.

SInce "Q_SLOTS " is #define Q_SLOTS I really do not know if
void start() is a slot or function .

Irrgadless of what it is - any suggestion HOW to trace "start" would be appreciated .

IMHO it is a function , so how do I find the "start" actual code , as used in
QBluetoothDeviceDiscoveryAgent class SOURCE CODE ?

Or can I run my app in debug mode to see how "start" interact with "blueZ"?

I am really looking for direct access to "blueZ" IF that is the underlining library used by Qt.
I need to see pass the Qt "wrappers".

[Pablo J. Rogina]

@AnneRanch said in Implementing bluetooth - general questions - part 3 - trace bluetooth inquiry in Qt:

any suggestion HOW to trace "start" would be appreciated .

You may want to enable categorized logging and these are all the possible logging categories pertaining Bluetooth:

qt.bluetooth
qt.bluetooth.android
qt.bluetooth.bluez
qt.bluetooth.winrt
qt.bluetooth.winrt.service.thread

[Pablo J. Rogina]

@AnneRanch said in Implementing bluetooth - general questions - part 3 - trace bluetooth inquiry in Qt:

any suggestion HOW to trace "start" would be appreciated .

You may want to enable categorized logging and these are all the possible logging categories pertaining Bluetooth:

qt.bluetooth
qt.bluetooth.android
qt.bluetooth.bluez
qt.bluetooth.winrt
qt.bluetooth.winrt.service.thread

[AnneRanch]

@Pablo-J-Rogina Thanks, I think this will do nicely and not only on bluetooth.
Appreciate your post.