以下程序的输出是什么？ 您如何跟踪这样的程序？

Question

I have to determine the output of the program below. The answer is BEADC , but i'm not sure how to trace the program. Here's is what i know:

We start off by declaring a structure called road_trip with data type trip . In the main program we assign values to the member place . Then a linked list is created where the starting node is s2 and ending node is s3 . Because you can see that the address of starting node is being stored in a separate pointer ptr which is often known as head pointer. s3 is the ending of the linked list because if you see the s3 you may notice that next of s3 is NULL . That means s3 is not referring any other node.

What I don't understand is how does the program prints the values stored in s2 ( B ), s5 ( E ), s1 ( A ), s4 ( D ), and s3 ( C ) in that order. I'm sure it has to do with the two lines I wrote comments after. An explanation would be of great help.

#include <stdio.h>

typedef struct road_trip
{
    char place;
    struct road_trip* next;
}trip;

int main (void)
{
    trip s1, s2, s3, s4, s5, s6;
    trip *ptr;

    s1.place = 'A';
    s2.place = 'B';
    s3.place = 'C';
    s4.place = 'D';
    s5.place = 'E';

    s5.next = &s1;
    ptr = &s2;
    s1.next = &s4;
    s3.next = NULL;
    s4.next = &s3;
    s2.next = &s5;

    while (ptr != NULL)
    {
        printf("%c ", ptr -> place);    /*I don't get this line*/
        ptr = ptr -> next;    /*I don't get this line*/
    }
    return 0;
}

Answer 1

 printf("%c ", ptr -> place); /*I don't get this line*/ ptr = ptr -> next; /*I don't get this line*/ 

• Here, ptr is a pointer to the trip (ie, struct road_trip ) .
• when you want to access the members of a structure using a pointer, then -> ( Structure dereference operator) operator is used

---

The answer is "BEADC", but i'm not sure how to trace the program.

s5.next = &s1;
ptr = &s2;
s1.next = &s4;
s3.next = NULL;
s4.next = &s3;
s2.next = &s5;

The above part of the code is useful to trace the program...

Here

• ptr points to s2

and:

• next member of s2 points to s5
• next member of s5 points to s1
• next member of s1 points to s4
• next member of s4 points to s3
• next member of s3 points to NULL

You've given the correct explanation in your question

---

What I don't understand is how does the program prints the values stored in s2 (B), s5 (E), s1 (A), s4 (D), and s3 (C) in that order

Now when you iterate using ptr this way :

while (ptr != NULL)
{
    printf("%c ", ptr -> place);
    ptr = ptr -> next;
}

Tracing the loop :

• on first iteration B gets printed because ptr points to s2 and thus ptr->place is same as s2.place (so using the Structure dereference operator the values of place member accessed)

• then ptr = ptr->next is same as ptr = s2->next and thus ptr = &s5 as s2->next points to s5

• Similarly, this loop continues and you get the output in the way their next members are connected

• The loop ends after printing s3 because s3->next points to NULL and thus ptr becomes =NULL after printing s3

---

Further reading :

• This is a typical implementation of linked lists (click to know more) .

• This type of list where one member of structure points to next structure of the list is known as singly linked list (click to know more) .

Answer 2

ptr is a pointer to s2, so displaying char ptr->place will display place of trip s2. Then ptr is being changed to next pointer - next of s2 is s5. Displaying ptr -> place will display now place of trip s5. Then ptr changes to next of s5, which is s1 and it goes untill it reaches s3, which next pointer is NULL.