Creating a generic linked list in c++ without stl -

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can creating generic linkedlist without stl. how declare head in main. struct node<>* head ? or struct node* head ? got error using both , template declaration cannot appear @ block scope

#include <iostream> using namespace std;   template<class t> struct node {     t data;     struct node<t>* next; };  template<class t> void push(struct node<t>** h,t dat) {     struct node<t> * newnode=(struct node<t> * )malloc(sizeof(struct node<t>)) ;     newnode->data=dat;     newnode->next=*h;     *h=newnode;  }   int main() {      struct node<>* head=null;     struct node<>* current;     int a=10;     float f=10.1;       push<int>(&head,a);     push<float>(&head,f);       current=head;     while(current)     {         cout<<current->data;         current=current->next;     }      //code     return 0; }

first of all, weird mix of c , c++ style programming. let's ignore , focus on real question. primary issue you're not specifying type parameter when referencing node (should node<t> when use it). changing first bit to:

template<class t> struct node {     t data;     struct node<t>* next;  };  template<class t> void push(struct node<t>** h,t dat) // <-- use node<t> everywhere {     struct node<t> * newnode=(struct node<t> * )malloc(sizeof(struct node<t>)) ;     newnode->data=dat;     newnode->next=*h;     *h=newnode;  }

should need go. there, you're referring node<t> everywhere in push. same apply main(). malloc work, node<t> have definite size.

that said, you'll find bit cleaner use node<t> *example = new node<t> , delete example instead.

there number of other improvements made move more c++ realm i'm focusing on direct question here; move on rest later.


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