CODEWITHSUNDEEP
c
Angus
Angus

Reputation: 12621

Segmentation fault in program using linked list

Code:

#include<stdio.h>
#include<malloc.h>
struct details{
  char *name;
  int no;
  struct details *info;
};

//ADDING THE LINKED LIST
void add(struct details **info,int no,char * name){
  struct details *temp=malloc(sizeof(struct details));
  temp = *info;
  if(temp == NULL){
      temp = malloc(sizeof(struct details));
  }
  else{
    if(temp->info == NULL){
       temp->info = malloc(sizeof(struct details));
       temp = temp->info;
    }
  }
  temp->no = no;
  temp->name = name;
}

//DISPLAYING THE LINKED LIST
void display(struct details *info){
  while(info!=NULL){
    printf("\nThe List is:\n","\n no: \tname:\n","%d","%s and link:%d",info->no,info->name,info->info);
    info = info->info;
  }
}

//MAIN PROGRAM
int main()
{
 struct details* ptr;
 char *name,ch;
 int no;
 int select_option;
 ptr = NULL;
 printf("\n   ***MAIN MENU***   \n1.Add Element \n2.Delete Element \n3.Search Element \n4.Linked List Concatenation \n5.Invert Linked List \n6.Diplay Elements \n Please Enter your choice:(eg:1,2,3,4,5,6)\n");
 scanf("%d",&select_option);
do{
 switch(select_option){
   case 1:
        printf("Enter no to add:");
        scanf("%d",&no);
        printf("Enter name to add:");
        scanf("%s",name);
        add(&ptr,no,name);
        break;
   case 6:
        display(ptr);
        break;
   default:
        printf("INVALID CHOICE!");
        break;
 }
 printf("Do u wish to continue?(y/n):");
 scanf("%c",&ch);
}while(ch == 'y' || ch == 'y');
 return 0;
}

I'm trying to write a simple program using a linked list to add and display data. But its throwing me a segmentation fault. I hope that I have initialized all the pointers with the memory. All help appreciated.

Upvotes: 0

Views: 486

Answers (4)

hmjd
hmjd

Reputation: 121961

This:

temp->name = name;

does not copy name to temp->name but assigns temp->name to the same address as name, which is local to the function. You need to malloc() and strcpy():

temp->name = malloc(strlen(name) + 1);
strcpy(temp->name, name);

Remember to free(temp->name); when no longer required.

Additionally (as pointed out by Luchian), when reading from stdin:

char *name;
...
scanf("%s",name);

name has no memory allocated to. Declare it as an array but you need to protect against writing beyond the end of it:

char name[128];
...
scanf("%127s",name);

Upvotes: 3

thumbmunkeys
thumbmunkeys

Reputation: 20764

You forgot to allocate name

     char *name

Upvotes: 1

Luchian Grigore
Luchian Grigore

Reputation: 258548

You haven't:

char *name;
///....
scanf("%s",name);

You never allocated memory for name.

You could do char name[50]; or whatever, but beware of overflows. Guard against them.

Upvotes: 2

MByD
MByD

Reputation: 137272

Both hmjd and Luchian Grigore answers are a prior problems, but also:

You don't initialize temp->info at any point after a malloc, so it is never NULL, but yet temp->info is an invalid address.

Either initialize it explicitly or use calloc, which initializes the allocated buffer, instead of malloc.

Upvotes: 1

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