Question

Assume entries in a linked list are of type struct listrec:

struct listrec

{

            char                 value;

            struct listrec    *next;

};

Write a main() routine in which you create the following linked list:

| a |   --|---->   | c |   --|-----> | w |   --|----> NULL

Write a function called printlist that takes a pointer to the start of a linked list and prints out all the nodes in the list in sequence. The inferface of this function is

void printlist(listrec *start_of_linked_list)

{

          // print out all the nodes in the list

          …

}

In your main( ), you may have

void main( )

{

            listrec *head_old;

            head_old = (beginning memory address of the linked list)

          printlist(head_old); // print out the information in the linked list

                        …

}

Answer 1

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

struct listrec {
    char value;
    struct listrec *next;
};

typedef struct listrec listrec;

void printlist(listrec *start_of_linked_list) {
    if(start_of_linked_list == NULL) {
        printf("NULL\n");
    } else {
        printf("| %c |   --|---->   ", start_of_linked_list->value);
        printlist(start_of_linked_list->next);
    }
}

int main() {

    listrec *head_old = NULL;
    head_old = (listrec *)malloc(sizeof(listrec));
    head_old->value = 'a';
    head_old->next = (listrec *)malloc(sizeof(listrec));
    head_old->next->value = 'c';
    head_old->next->next = (listrec *)malloc(sizeof(listrec));
    head_old->next->next->value = 'w';
    head_old->next->next->next = NULL;
    printlist(head_old); // print out the information in the linked list
    return 0;
}

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