C++ Programming - Objects and Classes - Discussion
Discussion Forum : Objects and Classes - Programs (Q.No. 1)
1.
What will be the output of the following program?
#include<iostream.h>
class Bix
{
public:
int x;
};
int main()
{
Bix *p = new Bix();
(*p).x = 10;
cout<< (*p).x << " " << p->x << " " ;
p->x = 20;
cout<< (*p).x << " " << p->x ;
return 0;
}Discussion:
4 comments Page 1 of 1.
Rahul said:
1 decade ago
@Venkatraman, shivya.
bix *p - represents pointer(p) to a class(bix) in the same way as int *p(i.e, pointer(p) to an int) is declared.
bix *p= new bix();
This is just to invoke bix class constructor with p as a pointer to that (bix) class type .
Now (*p).x is the same as p->x because both represents pointer to a type (i.e, in first case we have class bix and in second case it is taken as a structure ). Flashback the use of pointers in structures. Remember? no? Class is also a type of structure, the only difference is that we cannot have functions in structure and by default access specifier in structures is PUBLIC.
bix *p - represents pointer(p) to a class(bix) in the same way as int *p(i.e, pointer(p) to an int) is declared.
bix *p= new bix();
This is just to invoke bix class constructor with p as a pointer to that (bix) class type .
Now (*p).x is the same as p->x because both represents pointer to a type (i.e, in first case we have class bix and in second case it is taken as a structure ). Flashback the use of pointers in structures. Remember? no? Class is also a type of structure, the only difference is that we cannot have functions in structure and by default access specifier in structures is PUBLIC.
Nazmul said:
1 decade ago
Bix *p = new Bix(); in this *p is a pointer to the class not a regular object.
So whenever the value of the public variable changed within a class the pointer remains unchanged to that address.
So if the variable changes pointer will also take care of that.
p->x = 20; assigned by the object and still *p is pointed to that class.
So
cout<< (*p).x << " " << p->x ; will print 10 10 20 20.
So whenever the value of the public variable changed within a class the pointer remains unchanged to that address.
So if the variable changes pointer will also take care of that.
p->x = 20; assigned by the object and still *p is pointed to that class.
So
cout<< (*p).x << " " << p->x ; will print 10 10 20 20.
Venkatarama said:
1 decade ago
PLEASE GIVE A BETTER EXPLANATION on the:
cout<<(*p).x<<""<<p->x;
Why the output is 10 10 20 20. I am not understanding (*p).x.
Please give me difference on p->x and (*p).x.
cout<<(*p).x<<""<<p->x;
Why the output is 10 10 20 20. I am not understanding (*p).x.
Please give me difference on p->x and (*p).x.
Shivya said:
1 decade ago
Please explain it in detail.
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