pointers

[bbcc]

Greetings,

I haven't seen such a statement in c++ so far. Is G4Track a pointer?

void ScoreNewTrack(const G4Track*);

Thnak you

[onur]

G4Track is a type.
But const G4Track* is a pointer to a const G4Track.

So you will use this function like this:

G4Track myG4Track(/*provide the necessary arguments to create a G4Track*/);
ScoreNewTrack(& myG4Track); // & x yields the address of the variable x, i.e. create a pointer to x

// or

G4Track* myG4TrackPointer = new G4Track(/*provide the necessary arguments to create a G4Track*/);
ScoreNewTrack(myG4TrackPointer);

[bbcc]

So if it is a pointer, is there any difference of the two writings below?

void ScoreNewTrack(const *G4Track);

and

void ScoreNewTrack(const G4Track*);

[Moschops]

The first one is incorrect syntax. It is meaningless. It states that the function accepts an object named G4Track, of type const *, which is meaningless. You cannot have just a const pointer, it must be a const pointer to a type.

The second one states that the function accepts a pointer to a const G4Track, which makes a lot of sense.

[onur]

What you can switch (to some extend is the const):

void ScoreNewTrack(const G4Track*);
// is the same as
void ScoreNewTrack(G4Track const *);
// but NOT the same as
void ScoreNewTrack(G4Track * const);

Some prefer the second version because you can read it "from right to left":
Pointer to a const G4Track.
The third version is different since if you read it "right to left" again:
A const pointer to a G4Track.
Here the pointer is const and not the pointed to G4Track.

[bbcc]

So can I say that the pointer has a type but not a name?

The type is a const G4Track but there is no name for the pointer!

[Moschops]

That's correct. You do not need to name the parameter in the function prototype; the compiler has no use for them in the prototype. The actual function definition needs to name the parameters so they can be used in the function body.