What's arrow -> means in C++?

HelloWorld · #1 (**permalink**) 11-19-2007, 08:22 AM

What is -> means in C++...? I know that it's not the way to call a method right...?

thx

Bacteria · #2 (**permalink**) 11-19-2007, 10:50 AM

well when you want to call a function using pointers you usde ->

Example :
We have a class Student
and this class has the function getstudentname();

in the main :

Code:

void main()
{
      Student s;
      Student *sptr;
 
sptr=&s;// sptr is pointing to all the functions of s
 
so we use :
sptr->getstudentname();

to use the function of s

TeraTask · #3 (**permalink**) 11-19-2007, 10:57 AM

Bacteria: Please remember to use code tags.

akshay_pirate · #4 (**permalink**) 11-21-2007, 12:19 PM

well buddy,-> operator is used when you want to use data members or member functions of a class using that class's object's pointer.
ex:

class C{
public:
int x;
void get(int a){
x=a;
}
};
void main(){
C c;
C *c1;
c1=c;

c1->a(3);
cout<<c1->x;
}

HelloWorld · #5 (**permalink**) 11-21-2007, 12:57 PM

Quote:

Originally Posted by akshay_pirate

well buddy,-> operator is used when you want to use data members or member functions of a class using that class's object's pointer.
ex:

class C{
public:
int x;
void get(int a){
x=a;
}
};
void main(){
C c;
C *c1;
c1=c;

c1->a(3);
cout<<c1->x;
}

what's the difference with this then:

Code:

void main() {
 C c;
 c.a(3);
}

TeraTask · #6 (**permalink**) 11-21-2007, 01:17 PM

For the difference between the . and -> operators, I suggest checking out Member Access Operators: . and -> (C++) . Essentially, the difference has to do with whether you're dealing with a pointer or not and needing to dereference.

Quote:

The –> operator dereferences the pointer. Therefore, the expressions e–>member and (*e).member (where e represents a pointer) yield identical results (except when the operators –> or * are overloaded).

akshay_pirate · #7 (**permalink**) 11-28-2007, 06:26 AM

no difference,both do the same work.

Bench · #8 (**permalink**) 11-30-2007, 04:24 AM

Just to add to this discussion - the -> operator is a short-hand way of writing (*foo).x where 'foo' is a pointer to a class/struct/union type.

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