Understanding warning "Returning data of temporary object"

saa_

I have written a function to convert QString to char*

    char* qStringToCharPtr(QString string, int* length)
    {
        QByteArray byteArray = string.toLocal8Bit();
        *length = byteArray.length();
        return byteArray.data();
    }

And I receive the warning: "Returning data of temporary QByteArray". I understand that this is because the byteArray object is destroyed after the function call, and I am returning a pointer to the data stored in QByteArray .
Now change the function to this

    char* qStringToCharPtr(QString string, int* length)
    {
        QByteArray byteArray = string.toLocal8Bit();
        *length = byteArray.length();
        char * ch = byteArray.data();
        return ch;
    }

And with these changes, I don't get the warning. So my question is how come I don't receive the warning with the changes I made? All I did was to create a new pointer that points to the data in byteArray and return that instead?

mrjj

Hi
It tries to tell you that byteArray data is very temporary and
you return a pointer to its internal data.
In version 2, you hide that fact as
char * ch = byteArray.data();
is indeed valid for that line
and from "return ch; " it cannot know that anymore but the effect is still the same.

If we look in docs
https://doc.qt.io/qt-5/qbytearray.html#data
it says
"The pointer remains valid as long as the byte array isn't reallocated or destroyed. "

and QByteArray byteArray is a local variable so as soon as the function ends, it will be dellocated.

So its better to use .data() in place and not via a function.

saa_

That makes sense thanks. Suppose I need to use another object that is very temporary like QByteArray in a function and return some of its internal data, what would be the way to do it? Maybe it is a very niche situation, but still nice to know

mrjj

@saa_
Hi
Well it kinda depends on what we want to return.
Sometimes we can get away with simply returning a copy.

Like
QString SomeFunc()

then the Qstring does the copying for us.

if we are talking a char * from a container like qbytearry the best way would be to allocated storage for it
as docs shows.

QString tmp = "test";
QByteArray text = tmp.toLocal8Bit();
char *data = new char[text.size() + 1];
strcpy(data, text.data());

and use that buffer. but that also places the burden on the caller to remember to deallocate the buffer.
So when this i needed to do then using a smart pointer to hangle the char * buffer can be used.

However, its not 100% the whole truth

https://en.cppreference.com/w/cpp/language/lifetime
(Temporary object lifetime)

So you might get away with code belov as when part of the calling expressions, the compiler might keep it alive.

  char* qStringToCharPtr(QString string)
    {
        return  string.toLocal8Bit().data();
    }