[SOLVED]Use of Iterator to access a QList

Hi,
Here is the code:

Code:

---

#include <QDebug>
#include <QList>
 
class Node
{
public:
    int a;
    int b;
};
int main(int argc, char *argv[])
{
    //case1
    QList<int> list1;
    list1.append(10);
    list1.append(3);
    QList<int>::iterator i;
    for (i = list1.begin(); i != list1.end(); ++i)
        qDebug() << *i;
 
    //case2
    QList<Node> list2;
    Node no = {10, 3};
    list2.append(no);
    QList<Node>::iterator j;
    for (j = list2.begin(); j != list2.end(); ++j)
        //here ???
        ;
 
}

---

In case 2, how can i access the a and b from each node?
Thanks

Code:

---

*i.a;

---

But I would preferre the index based approach for lists...

Good work, asking with a minimal compilable program!

Code:

---

//case2
    QList<Node> list2;
    Node no = {11, 12};
    list2.append(no);
    QList<Node>::iterator j;
    for (j = list2.begin(); j != list2.end(); ++j)
    {
        qDebug() << (*j).a << (*j).b;
    }

---

HIH

Johannes

humm ... are you certain of this?

ahhh ... ok ...

Code:

---

qDebug() << (*j).a << (*j).b;

---

TANKS

Damn, of course the brackets...:p

Just wanted to apologize for the cross/double post.. but, yes, that's right. You need the brackets.

You should consider, using a constructor for your class, which initializes those properties. Or use a struct and plain old c. :->

Code:

---

#include <QApplication>
#include <QDebug>
#include <QList>
 
class Node
{
public:
    int a;
    int b;
};
 
int main(int argc, char *argv[])
{
    QApplication app(argc, argv);
    //case1
    QList<int> list1;
    list1.append(10);
    list1.append(3);
    QList<int>::iterator i;
    for (i = list1.begin(); i != list1.end(); ++i)
        qDebug() << *i;
 
    //case2
    QList<Node> list2;
    Node no = {11, 12};
    list2.append(no);
    QList<Node>::iterator j;
    for (j = list2.begin(); j != list2.end(); ++j)
    {
        qDebug() << (*j).a << (*j).b;
    }
    return 0;
}

---

Johannes

Again :-> I should type faster!

How about

Code:

---

qDebug() << j->a << j->b;

---

Quote:

---

Originally Posted by JohannesMunk

Again :-> I should type faster!

---

I'm a little slow today also!

Slow, but you contributed something new, anyhow :->

But as the iterator itself, is not a pointer type, that might seem strange..

It has its *-operator overloaded to return the item, and (*a). is equivalent to a-> .. compiler magic.

Well its a matter of taste, I guess.

Joh