In c++ what does -> mean?

Question

Possible Duplicate:
When do I use a dot, arrow, or double colon to refer to members of a class in C++?

MainWindow::MainWindow(QWidget *parent) :
    QMainWindow(parent),
    ui(new Ui::MainWindow)
{
    ui->setupUi(this);
}

In the code above, what does -> mean? Thanks.

That's a strangely basic question to ask from the middle of some advanced code. How did you get this far? I'd recommend reading a good book on C++ for a bit until you're familiar with all the fundamentals. — Kerrek SB, Nov 14 '11 at 18:04
@KerrekSB, my guess is it's either an example or template project of IDE. — Michael Krelin - hacker, Nov 14 '11 at 18:05
@Nanocom: When you put a -1 on this, Did you tell the OP what should be searched for or what book should be referred? or even pointed him/her in the right direction? Just an overzealous -1 helps no one. — Alok Save, Nov 14 '11 at 18:10
This is pretty much basic and You need to pick up a [good C++ book](http://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list) to learn it. — Alok Save, Nov 14 '11 at 18:11
What... no one here has worked with Qt and Qt's ide (Qt Creator)? This is the basic 'template' for a Qt Gui app. — g19fanatic, Nov 14 '11 at 18:18
Why the hell are people downvoting this? A smug sense of superiority? It's a legitimate question, not poorly worded and polite. The only reason I can think of is that the question may be a duplicate, but there is a close vote feature for that very reason. — Tamás Szelei, Nov 14 '11 at 18:23
I downvoted it because a google search for "C++ operators" led to me wikipedia which gave a high level description of the operator. By using its suggested name I immediately found http://www.cplusplus.com/doc/tutorial/structures/ which seems to show its use nicely, showing a lack of searching as well as the fact that it's a duplicate. — Mark B, Nov 14 '11 at 18:31
@MarkB, StackOverflow aims to collect the answers for questions, not to send away beginners yelling RTFM at them. He might not even know what to search for! — Tamás Szelei, Nov 14 '11 at 18:34
All I can say is that How to think like a computer scientist c++ edition really isn't a good book to be reading by itself. — , Nov 14 '11 at 20:18

score 7 · Accepted Answer · edited Nov 14 '11 at 18:30

It means the object on the left side of the arrow is a pointer to an instance of a class or structure, and the name on the right is a member of that class or structure.

In the context:

ui->setupUI(this);

it means that ui is a pointer to an instance of the class Ui::MainWindow and the setupUI(this) is an invocation of the setupUI member function of that class with the parameter this.

When you have an actual instance of a class (or structure) instead of a pointer, then you use the notation:

Ui::MainWindow non_ptr; // not a pointer
non_ptr.setupUI(this);

Strictly, you can also write:

(*ui).setupUI(this);

However, you will rightly be ostracized if you write code like that rather than using the -> arrow operator.

Toomai · Answer 2 · 2011-11-14T18:13:34.667

2

In most cases, x->y is the same as (*x).y. Things like iterators can overload -> to do a bit of bonus stuff but in general it's a shorthand.

edited Nov 14 '11 at 18:13

answered Nov 14 '11 at 18:03

Toomai

3,974
1
20
22

5

Sometimes. Unless it's not. (Operators can be overloaded in C++.) – Kerrek SB Nov 14 '11 at 18:05
2

I think that overloading these particular operators is bad form, so it should be true in most reasonable situations. – Toomai Nov 14 '11 at 18:06
1

@Toomai: These operators are necessarily overloaded for e.g. iterators and smart pointers. – Oliver Charlesworth Nov 14 '11 at 18:07
True, sometimes, but then you can say that about any operator that can be overloaded. – Michael Krelin - hacker Nov 14 '11 at 18:10
2

@OliCharlesworth, yes they are of course overloaded for iterators, but they are overloaded in a way that it's semantically very similar (iterators mimic pointers). If not, than that's a bad practice and should be avoided. – Tamás Szelei Nov 14 '11 at 18:27

score 2 · Answer 3 · answered Nov 14 '11 at 18:04

2

ui-> is equivalent to (*ui). (access the member of the object pointed to by ui).

answered Nov 14 '11 at 18:04

Michael Krelin - hacker

138,757
24
193
173

score 2 · Answer 4 · answered Nov 14 '11 at 18:10

2

ui->setupUi(this) is actually a short way of writing (*ui).setupUi(this).

P.S.: Somebody had downvoted your question, so I upvoted it. To the downvoter: come on! So what if this is a trivial question? It might be a perfectly reasonable one for a person who is just starting to learn C++.

answered Nov 14 '11 at 18:10

2

I did not downvote.But come on now.Who starts to learn `C++` with code like `MainWindow::MainWindow(QWidget *parent) : QMainWindow(parent), ui(new Ui::MainWindow)`?This is very scary for a beginner.... – Cratylus Nov 14 '11 at 18:12
True that. But we don't know his situation. Maybe it just so happened that he's stuck with some Qt code that he needs to fix real quick. – Nov 14 '11 at 18:20
2

`MainWindow::MainWindow(QWidget *parent) : QMainWindow(parent), ui(new Ui::MainWindow)` is exactly what will be generated by IDE on creation of a new Qt GUI application - probably the simplest starting point for a new user of Qt. – sam-w Nov 14 '11 at 18:27
That's auto-generated code that QtCreator generates. – Tamás Szelei Nov 14 '11 at 18:32

score 1 · Answer 5 · answered Nov 14 '11 at 18:04

1

It is a notation in both C and C++ to indicate that we access a variable via a pointer

answered Nov 14 '11 at 18:04

Cratylus

52,998
69
209
339

score 1 · Answer 6 · answered Nov 14 '11 at 18:12

Assuming ui is a pointer type (as implied by initializing it with new UI::MainWindow), then ui->setupUi is equivalent to *(ui).setUi.

However, it's also possible to overload operator-> for a class, in which case it's basically equivalent to ui.operator->(whatever). It's probably worth noting that the semantics of an overloaded operator-> are somewhat unusual: x-> is interpreted as (x.operator->())->y, so the overloaded operator-> must return some type for which returned_object->y is also legal -- either another object that also overloads operator->, or else an actual pointer.

In c++ what does -> mean?

6 Answers6