Appending pointers to QList

Question

I need to insert pointers of classes (inherited from QObject) into a QList. I know that the following syntax can be used:

.h

QList<MyObject*> list;

.cpp

list.append(new MyObject("first", 1));
list.append(new MyObject("second", 2));
...

and then free memory:

if(!list.isEmpty())
{
    qDeleteAll(list);
    list.clear();
}

This should be valid and does not cause any memory leaks (as far as I know). However, I need to initialize objects before adding them to the collection. Can the following piece of code cause some errors like memory leaks or dangling pointers (I'll use the same way to delete pointers as above)?

MyObject *obj;

for(i = 0; i < 5; i++)
{   
    obj = new MyObject();
    if(!obj.Init(i, map.values(i)))
    {
        // handle error
    }
    else
    {
        list.append(obj);
    }
}

Thanks.

Answer 1

Use RAII (Resource Allocation Is Initialization). Initialize the object in constructor directly.

Then the code would look like:

for(i = 0; i < 5; i++)
{   
    list.append( new MyObject( i, map.values(i)));
    // In case of initialization failure, throw exception from the constructor
}

Answer 2

if you take care of "obj" (the allocated but not initialized instance) in the "// handle error" case, your code is ok.

Answer 3

You can use QScopedPointer..

From the Qt 4.6 documentation,

The QScopedPointer class stores a pointer to a dynamically allocated object, and deletes it upon destruction. Managing heap allocated objects manually is hard and error prone, with the common result that code leaks memory and is hard to maintain. QScopedPointer is a small utility class that heavily simplifies this by assigning stack-based memory ownership to heap allocations, more generally called resource acquisition is initialization(RAII).

Hope it helps..

Edit:

For example,

You will use,

QScopedPointer<QWidget> p(new QWidget());

instead of

QWidget *p = new QWidget();

and add the QScopedPointer into your QList without worrying about memory leak and dangling pointers.