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passing different struct in a linked list using template

1 New Member
Hi dis my codig..
hav a luk..

#include<iostre am>

#include<cstrin g>

using namespace std;

template<class T>class List{

struct node{

node *link;

T no;

node(T t):no(t){}

};

node *head;

public:

List(){head->link=NULL;}

void print_all()

{

node *p;

for(p=head;p!=N ULL;p=p->link)

cout<<p->no;

}

void add_one(T t)

{

node *p;

if(head->link==NULL)

{

head=new node(t);

head->link=NULL;

return;

}

for(p=head;p->link!=NULL;p =p->link);

node *temp;

temp=new node(t);

temp->link=NULL;

p->link=temp;

}

};

class employee

{

char name[20];

int emp_no;

int salary;

public:

employee(char s[20],int t,int t2)

{

strcpy(name,s);

emp_no=t;

salary=t2;

}

};

void main()

{

employee e1("ruby",20,10 00);

List <employee> e;

e.add_one(e1);

e.print_all();

}

//error C2679: binary '<<' : no operator found which takes a right-hand operand of type 'employee' (or there is no acceptable conversion)

i got the above error..
is this coding correct or not..
if it is wrong please tell the correct coding..
Mar 26 '08 #1
1 1760
Banfa
9,065 Recognized Expert Moderator Expert
The code is wrong or incomplete, in your function display_all you have

Expand|Select|Wrap|Line Numbers
  1. cout<<p->no;
  2.  
But the type of p->no is T from the template so when you instantiated the list with T as a structure the compiler no longer knows how to call cout with that structure.

You can solve this by

1. Removing display_all

2. Extending cout by writing a << operator that takes cout and T as it's parameters and outputs your structure.
Mar 26 '08 #2

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