Self referential Structure - c++

Question

I am writing 2 codes for self referential structures :
1st:
struct node
{
int a;
struct node *link;
};
int main()
{
struct node n;
n.a = 5;
cout<< n.a << "\t" ;
cout << n.link ;
return 0;
}
Output : ̀ 5 0x40185b`
2nd:
struct node{
int a;
struct node *link;
};
int main(){
struct node n;
n.a = 5;
cout << n.a << "\t";
cout << *n.link ;
return 0;
}
Output: error:link was not declared in this scope.
Please tell me what is really happening in the code??
Why does a garbage value is thrown??
How can I initialize a self referential structure pointer??

Answer 1

I suppose that's what you want to do:
int main(){
struct node n;
n.a = 5;
n.link = NULL; // initialize the link
cout << n.a << "\t";
cout << n.link;
return 0;
}
*(n.link) will only be valid if n.link points to a valid node object.
And cout << *(n.link); will only be valid if you declare an operator<< for node (cout << n.link; is valid as it outputs the adress, not the value).
For instance, this would work much better:
#include <iostream>
struct node{
int a;
node *link; // Note: no need to prefix with struct
};
std::ostream& operator<<( std::ostream& str, const struct node& n )
{
str << n.a << "\t -> ";
if ( n.link )
str << *n.link;
else
str << "NULL";
return str;
}
int main(){
node n1; // Note: no need to prefix with struct
node n2; // Note: no need to prefix with struct
n1.a = 5;
n1.link = &n2;
n2.a = 6;
n2.link = NULL;
cout << n1;
return 0;
}
It outputs 5 -> 6 -> NULL