CODEWITHSUNDEEP
cfree
Kashyap Sripathi
Kashyap Sripathi

Reputation: 1

how does the free() function work?

#include <stdio.h>
#include <stdlib.h>
struct node{
int data;
    struct node* next;
};

struct node* nc(int data) {

    struct node* temp = (struct node*)(malloc(sizeof(struct node)));
    temp->data = data;
    temp->next = NULL;
    return temp;
}
int getcount(struct node* head){
    int count = 0;
    struct node* temp= head;
    while(temp!=NULL){
        count++;
        temp=temp->next;
    }
    return count;
}

int getpositionofelement(int data,struct node* head){
    int count=0;
    while(head->data!=data){
        count++;
        head=head->next;
    }
    return count;
}
int main()
{
    struct node h;
    struct node* head = &h; //(struct node*)malloc(sizeof(struct node));
    h.data=5;
    struct node* a = nc(19);
    struct node* b = nc(25);
    struct node* c = nc(12);
    h.next = a;
    a->next=b;
    b->next=c;
    free(b);
    int count = getcount(head);
    printf("the count is %d \n",count);
    int l = 5;
    printf("the position of %d in linkedlist is %d \n",l,getpositionofelement(l,head));
    while(head!=NULL){
        printf("%d \n",head->data);
        head=head->next;    
    }
}

When I free the node b why isn't the linked list getting terminated at b? How does the free() function work after taking a pointer as argument?

Upvotes: 0

Views: 94

Answers (1)

nicomp
nicomp

Reputation: 4647

free() deallocates memory from the heap, that's all. You need to set your pointer to null. Your pointer can contain an address to memory that's alreay been free'd: it 's an invalid address but it's still store-able in your pointer variable.

Upvotes: 1

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