如何计算链表中的节点数？为什么输出显示节点数为“2”？

Question

I have written ac program to find the number of nodes in a link list. But the problem arises because the value of count that I am printing comes out to be '2'.

My exact output looks like->

NUMBER OF NODES ARE 2

What am I doing wrong here?

//the code for the program is here:-

#include<stdio.h>
#include<stdlib.h>

struct node
{
 int data;
 struct node *next;
}*first=NULL;

void create(int a[],int n)
{
 struct node *t,*last;
 first=(struct node *)malloc(sizeof(struct node));
 first->data=a[0];
 first->next=0;
 last=first;
 int i;
 for(i=1;i<n;i++)
 {
  t=(struct node *)malloc(sizeof(struct node));
  t->data=a[i];
  t->next=NULL;
  last->next=t;
  last=first;
 }
}

void count(struct node *p) //function to count number of nodes
{
 int count=0;
 while(p!=NULL)
 {
  count++;
  p=p->next;
 }
 printf("Number of nodes are %d ",count);
}

int main()
{
 int a[]={1,2,3,4};
 create(a,4);
 count(first);
 return 0;
}

Answer 1

I think that in the loop within the function create

for(i=1;i<n;i++)

you mean

last = last->next;

instead of

last=first;

Pay attention to that it is a bad idea to declare the pointer to the head node as global and when functions depend on global variables.

Also the user can pass to the function the number of elements of an array equal to 0. Also allocation of memory can fail.

I would declare the function the following way

size_t create( struct node **head, const int a[], size_t n )
{
    // if the list is not empty free its nodes
    while ( *head != NULL )
    {
        struct node *current = *head;
        *head = ( *head )->next;
        free( current );
    }

    size_t i = 0;

    for ( ; i < n && ( *head = malloc( sizeof( struct node ) ) ) != NULL; i++ )
    {
        ( *head )->data = a[i];
        ( *head )->next = NULL;

        head = &( *head )->next;
    }

    return i;
}

And call the function like

size_t n = create( &first, a, sizeof( a ) / sizeof( *a ) );

In this case the function returns the number of created nodes in the list.

Here is a demonstrative program.

#include <stdio.h>
#include <stdlib.h>

struct node
{
    int data;
    struct node *next;
};

size_t create( struct node **head, const int a[], size_t n )
{
    // if the list is not empty free its nodes
    while ( *head != NULL )
    {
        struct node *current = *head;
        *head = ( *head )->next;
        free( current );
    }

    size_t i = 0;

    for ( ; i < n && ( *head = malloc( sizeof( struct node ) ) ) != NULL; i++ )
    {
        ( *head )->data = a[i];
        ( *head )->next = NULL;

        head = &( *head )->next;
    }

    return i;
}

void output( const struct node *head )
{
    for ( ; head != NULL; head = head->next )
    {
        printf( "%d -> ", head->data );
    }

    puts( "null" );
}

int main(void) 
{
    struct node *head = NULL;

    int a[] = { 1, 2, 3, 4 };
    const size_t N = sizeof( a ) / sizeof( *a );

    size_t n = create( &head, a, N );

    printf( "There are %zu nodes in the list\n", n );

    printf( "They are " );

    output( head );

    return 0;
}

Its output is

There are 4 nodes in the list
They are 1 -> 2 -> 3 -> 4 -> null