How to add a QString value to QList of QLists?

Eduardo12l

I've tried this:

QList<QList<QString>>Temporales
QList<QString> DeFNombre
QList<QString> DefGruop
QString aux = DeFNombre[0]+DefGruop[0];
Temporle[0].append(aux)

However, this way crash when debugging. What can i do?

koahnig

@Eduardo12l

Are you aware that there is a QStringList in Qt?

QStringList may be extended using the operator<<.
The classical push_back is another option.

Where is it exactly crashing when you do it this way?

When using the five lines of code given above, it will crash of course, because DeFNombre and DefGruop are still empty.

Eduardo12l

@koahnig

It means that i have no alternative and I must use QStringList ?
Those both QList are not empty they were initialized before, the only one that is empty is QList of QLists - Temporales

Sorry i forgot it. It crashed at :

Temporales[0].append(aux)

koahnig

@Eduardo12l

What do you mean by crashed?

A crash is typically when an crashes (stops) during execution with some stack output.

What you are seeing is compile error and not a crash. In such cases the error message tells you what is wrong.

In your first post you show:
@Eduardo12l said in How to add a QString value to QList of QLists?:

QList<QList<QString>>Temporales
QList<QString> DeFNombre
QList<QString> DefGruop
QString aux = DeFNombre[0]+DefGruop[0];
Temporle[0].append(aux)

which cannot be compiled without a compile error.

In your second post, it shows at least the problem, but it is still not a complete code line, since it will not compile as it is shown.

@Eduardo12l said in How to add a QString value to QList of QLists?:

@koahnig

It means that i have no alternative and I must use QStringList ?

Those both QList are not empty they were initialized before, the only one that is empty is QList of QLists - Temporales

Sorry i forgot it. It crashed at :
Temporales[0].append(aux)

In case those lines are only missing the semicolon in post and Temporales has already a first element, it should work.
See operator[]

koahnig

@Eduardo12l said in How to add a QString value to QList of QLists?:

@koahnig

It means that i have no alternative and I must use QStringList ?

No, it is not mandatory, but makes life a bit easier.

Eduardo12l

@koahnig When i run my app it suddenly stops and Qt Output tells me:
The program has unexpectedly finished.
C:\Qt\Saves\build-Proyecto1-Desktop_Qt_5_8_0_MSVC2015_64bit-Release\release\Proyecto1.exe crashed.

Instead, if I write this: (Instead of my first way)

Temporales.append(DefNombre); //DefNombre is a QList<QString> that is filled
 qDebug()<<Temporales.at(0).at(0);

I have no problems, i get from qDebug() the first column of the first row but it misses one value that comes from DefGrupo for that reason i cannot use this way
In other words my main problem is that QList<QList<QString>> doesnt let me to append QStrings

Eduardo12l

This post is deleted!

VRonin

From http://doc.qt.io/qt-5/qlist.html#operator-5b-5d:

Returns the item at index position i as a modifiable reference. i must be a valid index position in the list (i.e., 0 <= i < size())

Temporle[0] returns an invalid reference so it crashes. you have to insert an item before accessing it.


Temporle.append(QList<QString>());
Temporle[0].append(aux);

P.S.
QStringList inherits QList<QString> and just adds a few methods so they are basically the same thing

Taz742

@Eduardo12l
Used Qvector<>

    const int MAXN = 10;

    QVector<QVector<QString> > lst;

    lst.resize(MAXN);

    lst[0].push_back("foo");

    qDebug() << lst[0][0];

or

    const int MAXN = 10;

    std::vector<std::vector<std::string> > lst(MAXN);

    lst.at(0).push_back("faf");

    qDebug() << QString::fromStdString(lst[0][0]);

Taz742

or
@VRonin said in How to add a QString value to QList of QLists?:

Temporle.append(QList<QString>());
Temporle[0].append(aux);

QList<QList<QString> > lst;

for(int i = 0; i < 10; i++){
    lst.append(QList<QString>());
}

lst[0].append("foo");

qDebug() << lst[0][0];

captainBee

Qlist<T>*qlt;
for(int j; j<qlt.count(); j++) qlr.replace(j,"whatever");