Extracting const char* from QStringList

[SPlatten]

I start off with a QString:

QString appName = QApplication::applicationName();

After the assignment appName contains "MillikanUI", I've verified this in the debugger. I then split this into a QStringList:

QStringList appNameList = appName.split("U");

appNameList now contains 2 elements:

[0] "Millikan"
[1] "I"

I then try to convert the first element to a const char*:

const char* baseName = qPrintable(appNameList.at(APP_NAME_BASE));

APP_NAME_BASE is an enumerated type with the value 0. I'm using qPrintable because I tried toLatin1() and it returned the same sort of random characters that I'm now seeing returned by qPrintable. I've also tried appNameList[APP_NAME_BASE] with the same random results.

After the assignment baseName points to what looks like completely random content, nothing like the content of the element at 0 in the QStringList. What is going on?

[J.Hilk]

@SPlatten
the trick is to not use a temporary QByteArray, when you try to access its internal const char* array :D

int main(int argc, char *argv[])
{
    QApplication app(argc, argv);

    app.setApplicationName("MillikanUI");
    QString appName = QApplication::applicationName();
    QStringList appNameList = appName.split("U");

    //Non Temporary QByteArray
    QByteArray nonTemporayByteArray = appNameList.first().toLatin1();
    const char *data = nonTemporayByteArray.constData();

    qDebug() << data;

    //Temporary QByteArray
    data = appNameList.first().toLatin1().constData();

    qDebug() << data;

    return app.exec();
}

[JKSH]

@SPlatten said in Extracting const char* from QStringList:

After the assignment baseName points to what looks like completely random content, nothing like the content of the element at 0 in the QStringList. What is going on?

What's going on is described in the qPrintable() documentation https://doc.qt.io/qt-5/qtglobal.html#qPrintable : The char pointer will be invalid after the statement in which qPrintable() is used. This is because the array returned by QString::toLocal8Bit() will fall out of scope.

A const char * variable stores a single pointer. It does not store a string. You must ensure the memory that stores your pointed data remains valid for the lifetime of the pointer.

[SPlatten]

@JKSH , I also tried:

const char* baseName = ((appNameList[APP_NAME_BASE]).toLatin1()).data();

That didn't work either, what does work is:

QString appName(appNameList[APP_NAME_BASE]);
QByteArray appNameBA = appName.toLatin1();
const char* baseName = appNameBA.constData();

[JKSH]

@SPlatten said in Extracting const char* from QStringList:

That didn't work either

That's because you triggered the same phenomenon in both cases: A dangling pointer.

This is an important concept in C/C++. Take some time to ensure you intimately understand how dangling pointers come about; this will spare you lots headaches in the future.

[KroMignon]

@SPlatten said in Extracting const char* from QStringList:

const char* baseName = ((appNameList[APP_NAME_BASE]).toLatin1()).data();

This can not work!
appNameList[APP_NAME_BASE]).toLatin1() will create a new QByteArray instance, which will be directly destroyed after the call.
So baseName becomes a dangling pointer.

If you want this to work, you must ensure object instance will not be destroyed as long as you are working with it.

const auto buffer = appNameList[APP_NAME_BASE].toLatin1();
const char* baseName = buffer.constData();

baseName will be valid as longs as buffer is not destroyed.