UB or not

DmitryTS

I had an interesting question and confusion in my head, const QList<qlonglong>& keys = storeTypeChat.keys(); (storeTypeChat - QList) why there will be no dangling reference?

jsulm

@DmitryTS QList has no keys() method. Do you mean QMap/QHash?

Christian Ehrlicher

@DmitryTS said in UB or not:

why there will be no dangling reference?

Because the function returns a value which gets destroyed when the constant ref gets destroyed.

DmitryTS

@jsulm sorry, yes, storeTypeChat - QMap

DmitryTS

@Christian-Ehrlicher then it turns out that this is still a dangling ref?

JonB

@DmitryTS
Define exactly what you mean/think of "dangling reference" here.
Given @Christian-Ehrlicher's response of

Because the function returns a value which gets destroyed when the constant ref gets destroyed.

DmitryTS

@JonB will a constant reference extend the life of this temporary object or not?

JonB

You mean from the point of the function call till the end of the reference's/function's lifetime? Yes! Rather than my picking a reference for you, Googling c++ constant reference extend the life of this temporary object or not will give you various confirmatory links.

I believe you are using an lvalue here (const QList<qlonglong>& keys = ...), and for that you must use const for lifetime extension, apparently.

Christian Ehrlicher

@DmitryTS said in UB or not:

a constant reference extend the life of this temporary object or not?

I already answered it...