[SOLVED]Use of Iterator to access a QList

[graciano]

Hi,
Here is the 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

[Lykurg]

*i.a;But I would preferre the index based approach for lists...

[JohannesMunk]

Good work, asking with a minimal compilable program!



//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

[graciano]

humm ... are you certain of this?

[graciano]

ahhh ... ok ...


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


TANKS

[Lykurg]

Damn, of course the brackets...:p

[JohannesMunk]

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. :->



#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

[JohannesMunk]

Again :-> I should type faster!

[norobro]

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

[norobro]

Again :-> I should type faster!

I'm a little slow today also!

[JohannesMunk]

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