QT - Understanding following lambda expression for a SLOT

Question

I am currently trying to understand the new QT5 signal/slot syntax

connect(sender, &Sender::valueChanged, [=](const QString &newValue) {
       receiver->updateValue("senderValue", newValue);
   });

Now my question is where is the address of the receiver SLOT in the above expression ? I wanted to know this because what happens if a signal is in threadA and the slot is in thread B and I wanted it to be a queued connection ?

Answer 1

A slot is a piece of code, it doesn't "live" in a thread - a thread might run it or not, but the code itself doesn't belong to any thread. (If the slot is a member function, then the Qt object defined as the receiver belongs to a Qt thread - that's a property of the object, not the function.)

In the code you have above, the compiler generates an object that:

• captures receiver by value ([=])
• has a function-call operator that can be called with a reference to a const QString.

That object is passed to connect along with the other two arguments. It's not a QObject, so it doesn't have an owning thread in the Qt sense. What you need to make sure of is that:

• what receiver points to stays alive for as long as that signal is connected
• receiver->updateValue(...) is thread-safe - it will be called in sender's context/thread.

If receiver->updateValue needs to be called in receiver's thread/context, then do not use that syntax for the connect call, use the one where you specify both sender and receiver, and the connection type.