Creating linked list of strings

问题

I am new to data structures as well as to linked lists. I am working on project named as Amazon product availability checker using tree in C. So I want to store strings in each node of the tree but while storing strings the code is not showing any error but the output is also not getting printed. I have pass the node to print function to print the string but nothing is getting printed.

I have shared the code for one string and one node only. I am working on ubuntu and I am coding in the C language.

Here is my code:

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

typedef struct node {
    char clothing[10];
    struct node *next;
} node;

// 1) creating node

node *create(char const ch[]) {
    int i;
    node *head = NULL;
    node *temp = NULL;
    node *p = NULL;
    temp = (struct node *)malloc(sizeof(struct node));
    temp->clothing[10] = ch[10];
    temp->next = NULL;
    return temp;
}

// 2) print 

void print(node *head) {
    node *p = head;
    while (p != NULL) {
        printf("%s", p->clothing);
        p = p->next;
    }
}

int main() {
    node *head = create("clothing");
    print(head);
}

回答1:

Your create function is incorrect:

you do not test for potential malloc failure
you do not copy the string, but merely cause undefined behavior by attempting to write to clothing[10] which is beyond the end of the array. By the way, you read ch[10] which may well be out of bounds too. You should instead copy the string, while avoiding a buffer overflow if ch is too long.

Here is an improved version:

#incude <string.h>
#incude <stdlib.h>

node *create(char const ch[]) {
    temp = malloc(sizeof(node));
    if (temp != NULL) {
        temp->next = NULL;
        temp->clothing[0] = '\0';
        strncat(temp->clothing, ch, sizeof(temp->clothing) - 1);
    }
    return temp;
}

Since C99, there is a way to allocate a copy of the string without a limitation on its size, and without the need for a separate allocation and a pointer in the node structure. It is called a flexible array. Here is how it works:

typedef struct node {
    struct node *next;
    char clothing[];
} node;

node *create(char const ch[]) {
    size_t size = strlen(ch) + 1;
    temp = malloc(sizeof(node) + size);
    if (temp != NULL) {
        temp->next = NULL;
        memcpy(temp->clothing, ch, size);
    }
    return temp;
}

回答2:

node *addnode(node *after, const char *str)
{
    node *nptr;

    nptr = malloc(sizeof(*nptr));
    nptr -> partstr = malloc(strlen(str) + 1);
    /* error checking you need to add after every malloc */
    strcpy(nptr -> partstr, str);
    if(!after)
    {
        nptr -> prev = NULL;
        nptr -> next = NULL;
    }
    else
    {
        after -> next -> prev = nptr;
        nptr -> next = after -> next;
        after -> next = nptr;
        nptr -> prev = after;
    }
    return nptr;
}

回答3:

I have pass the node to print function to print the string but nothing is getting printed.

doing

temp -> clothing[10] = ch[10];

you write (and may be read) one character out of the string, the max index in temp -> clothing is 9

you want something like

strcpy(temp -> clothing, ch);

but take care to not go outside the field clothing because ch is too long

so can be

strncpy(temp -> clothing, ch, sizeof(temp -> clothing) - 1);
temp -> clothing[sizeof(temp -> clothing) - 1] = 0; /* useful if strlen(ch) >= 10 */

are you sure you do not want to replace char clothing[10]; by char * clothing; to not have that limit to 10 ?

来源：https://stackoverflow.com/questions/55550666/creating-linked-list-of-strings

标签

string

linked-list