制作链表时的困惑

Question

i am trying to understand a code of creating and displaying a linked list

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

int create(int n);
void print();

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

int main(){
    int n,data;
    printf("Enter the Number of Nodes\n");
    scanf("%d",&n);
    create(n);
    print();
}

int create(int n){
    int i=1,data;
    struct node *temp=(struct node*)malloc(sizeof(struct node));
    printf("Enter element %d:\n",i++);
    scanf("%d",&data);
    temp->data=data;
    temp->next=NULL;
    head=temp;
    while(n-->1){
        struct node *t=(struct node *)malloc(sizeof(struct node));
        printf("Enter element %d:\n",i++);
        scanf("%d",&data);
        t->data=data;
        t->next=NULL;
        temp->next=t;
        temp=t;
    }
    printf("Done :)\n");
}

void print(){
    struct node *temp=head;
    printf("Elements in list are:\n");
    if(temp==NULL)
        printf("List is Empty\n");
    else
    while(temp!=NULL){
        printf("%d ",temp->data);
        temp=temp->next;
    }
    printf("\n");
}

Now, i understand most of the working but there is some confusion I wanted to clarify.

The first node is created as head.the second node as t, which is then connected to head by pointing the next of head to itself. then how does the third node know that we are connecting it to second node(because all nodes are named t). is it because of temp = t at the end of the while loop, so at next run the new t node gets connected to the temp which is previous t node.

If that is the case, then I assume that only addresses are being connected to each other.

More precisely, with malloc we assign a memory of data and next to temp(which is pointer to node) then put first element in data and NULL in next and that is our first node. after that we have a head pointer which points to this first node.

My real confusion is what is happening is that we create new address bunches with each loop run and the print() uses those addresses to iterate from head to the last node. and exept head there is nothing to access our linked list after we exit the function.

CORRECT?

the terminology may not be to the point.

Answer 1

is it because of temp = t at the end of the while loop, so at next run the new t node gets connected to the temp which is previous t node.

Yes. temp keeps track of the latest node.

struct node *next holds the memory location of the next node.

and except head there is nothing to access our linked list after we exit the function.

Yes. Without knowing head , you wont be able to go through your list.

Answer 2

第三个节点通过语句 temp->next=t 连接到第二个节点（temp 的地址为第二个，'t' 是一个新节点），最后，你正在使 temp=t 所以在下一次迭代中，temp将有第三个节点地址等等。