Thread: Problem in Adjancey list

  1. #1
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    Problem in Adjancey list

    I have made program for an adjancy list. First column is dynamically allocated and it's next nodes are allocated through linked list. I am not getting proper output.Pls help

    Code:
    #include<stdio.h>
    #include<conio.h>
    #include<malloc.h>
    
    struct node
    {
    int info;
    struct node *next;
    }*temp;
    
    struct fcol
    {
    int info;
    struct node *next;
    }*firstcol;
    void createlist(int*,int);
    
    int main()
    {
    int *a,num,i,j;
    clrscr();
    printf("Enter the number of nodes...");
    scanf("%d",&num);
    
    a=(int *)malloc(num*num*sizeof(int));
    
    printf("Enter the adjacency matrix....");
    for(i=0;i<=num-1;i++)
    {
      for(j=0;j<=num-1;j++)
          scanf("%d",&a[i*num+j]);
    }
    
    createlist(a,num);
    for(i=0;i<=num-1;i++)
    {
      printf("node%d",firstcol[i].info);
      for(temp=firstcol[i].next;temp!=NULL;temp=temp->next)
         printf("->node%d",temp->info);
      printf("\n");
    }
    getch();
    return(0);
    }
    
    
    void createlist(int a[],int num)
    {
    int *row,i,j;
    temp=NULL;
    firstcol=NULL;
    //allocates 1d array for first col
    firstcol=(struct fcol*)malloc(num*sizeof(struct fcol));
    
    for(i=0;i<=num-1;i++)
    {
      firstcol[i].info=i+1;
      firstcol[i].next=NULL;
    }
    
    for(i=0;i<=num-1;i++)
    {
      for(j=0;j<=num-1;j++)
      {
        if(a[i*num+j]==1)
        {
           temp=firstcol[i].next;
           while(temp->next!=NULL)
    	 temp=temp->next;
    
           temp->next=(struct node*)malloc(sizeof(struct node));
           temp=temp->next;
           temp->info=j+1;
           printf("\t%d",temp->info);
           temp->next=NULL;
        }
      }
    }
    }

  2. #2
    the hat of redundancy hat nvoigt's Avatar
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    I am not getting proper output.Pls help
    My advice would be to change your program, so it gives "proper" output.






    You could also ask a smart question ( see sig ).
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  3. #3
    Registered User cbastard's Avatar
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    Code:
      firstcol[i].next=NULL;
      temp=firstcol[i].next;
     while(temp->next!=NULL)
    temp->next=(struct node*)malloc(sizeof(struct node));
    Now can you see some thing.

    some others.
    Code:
    a=(int *)malloc(num*num*sizeof(int));
    Casting malloc.read faq.

    Code:
    void createlist(int*,int);
    remember array are not pointers although this is not giving error
    Long time no C. I need to learn the language again.
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  4. #4
    Frequently Quite Prolix dwks's Avatar
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    Code:
    #include <malloc.h>
    Most compilers don't need that header -- if yours does, get a new compiler.

    Code:
    void createlist(int*,int);
    
    void createlist(int a[],int num)
    As cbastard said, your function prototype and definitions are different.

    Why do you have a struct fcol?
    Code:
    struct node
    {
    int info;
    struct node *next;
    }*temp;
    
    struct fcol
    {
    int info;
    struct node *next;
    }*firstcol;
    ->
    Code:
    struct node
    {
    int info;
    struct node *next;
    } *firstcol, *temp;
    Unless you're planning to make them different.

    And you're casting malloc. And you're not freeing your malloc()ed data.

    I am not getting proper output.
    What output are you getting, then? And what is "proper" output?
    dwk

    Seek and ye shall find. quaere et invenies.

    "Simplicity does not precede complexity, but follows it." -- Alan Perlis
    "Testing can only prove the presence of bugs, not their absence." -- Edsger Dijkstra
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  5. #5
    ATH0 quzah's Avatar
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    Quote Originally Posted by dwks
    Code:
    void createlist(int*,int);
    
    void createlist(int a[],int num)
    As cbastard said, your function prototype and definitions are different.
    Actually, it's not really true. Arrays degrade into a pointer to their first element when passed. Any time you pass a single dimension array, it degrades to a pointer of that type. Thus, passing an "int[]", ends up a "int*". So it really doesn't matter in this case. After all, this is perfectly legal:
    Code:
    int array[] = { 1, 2, 3 };
    int *ptr = array;
    
    ptr[0] = 5; /* <--- Perfectly fine to do. */
    Quzah.
    Hope is the first step on the road to disappointment.

  6. #6
    Frequently Quite Prolix dwks's Avatar
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    Okay, thanks. I always knew they had the same result, but I didn't know they were compatible.

    Well, at least you can't say that vaibhav is freeing his memory.
    dwk

    Seek and ye shall find. quaere et invenies.

    "Simplicity does not precede complexity, but follows it." -- Alan Perlis
    "Testing can only prove the presence of bugs, not their absence." -- Edsger Dijkstra
    "The only real mistake is the one from which we learn nothing." -- John Powell


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