链接的字符串列表C / Struct

Question

I'm to create a LinkedList that has functionalities of adding and printing strings within.

Attaching my code below.

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

struct Line {
       char tekst[20];
       struct Line *next;
};

void print(const struct Line* n);
void add(struct Line* n, char t[20], struct Line* next);

int main(void) {
    struct Line lin, *tekst;
    tekst = NULL;
    add(&lin, "ExampleString1", tekst);
    add(&lin, "ExampleString2", tekst);
    print(&lin);

    getch();
    return 0;  
};

void print(const struct Line* n) {
     for ( ; n; n = n->next )
         printf("%s", n->tekst);
     printf("\n");
}

void add(struct Line* n, char t[20], struct Line* next) {
     strcpy(n->tekst,t); // before: n->tekst[20] = t[20];
     n->next = next;
}

It writes some random numbers on the standard output, and then crashes the commandline. I've no idea whether tekst2[20] should even be here (I'm not sure how to call my function arguments here).

My goal is to make a list of strings then be able to add and print them.

Answer 1

I'm nearly certain this is what you're trying to do:

void add(struct Line** pp, const char *t)
{
    struct Line *p = malloc(sizeof(*p));
    strcpy(p->tekst, t);
    p->tekst2[0] = 0;
    p->next = NULL;

    while (*pp)
        pp = &(*pp)->next;
    *pp = p;
}

Your print() function is using the wrong format specifier for printing a string:

void print(const struct Line* n) 
{
     for ( ; n; n = n->next )
         printf("%s\n", n->tekst);
     printf("\n");
}

Putting it together in main() :

int main(void)
{
    struct Line *lst = NULL;

    add(&lst, "SomeTxt");
    add(&lst, "SomeMoreTxt");
    add(&lst, "YetMoreTxt");

    print(lst);

    getch();
    return 0;  
};

Output

SomeTxt
SomeMoreTxt
YetMoreTxt

I leave the proper list cleanup code as well as proper error checking as a task for you.

How It Works

This function utilizes a "pointer-to-pointer" idiom. When you pass &lst from main() , you're passing the address of a pointer variable. Pointers are nothing more than variables that hold addresses to stuff (of the type the pointer it declared, of course). Example:

int a;
int *b = &a;

declares a pointer-to- int , and assigns the address of an int to store within it. On the other hand, this:

int a;
int *b = &a;
int **c = &b;

declares what we had before, but c is declared to be a pointer-to-pointer-to-int . Just like we stored the address of an int in b , notice how we store the address of a int* in c . this is a very powerful feature of the language.

That said, the code works like this:

void add(struct Line** pp, const char *t)
{
    // node allocation stuff. nothing special here
    struct Line *p = malloc(sizeof(*p));
    strcpy(p->tekst, t);
    p->tekst2[0] = 0;
    p->next = NULL;

    // look at the pointer addressed by our pointer-to-pointer pp
    //  while it is not null, store the address of the `next` pointer 
    //  of that node in pp and loop.
    while (*pp)
        pp = &(*pp)->next;

    // pp now holds the address of the pointer we want to set with our new node
    //  it could be the address of the original pointer passed in (if the list was
    //  empty). or the address of some `next` member in the list. We really don't 
    //  care which. All we care about it is it addresses the pointer we need to 
    //  assign our new allocation to, so that is what we do.
    *pp = p;
}

Spend some time on the net researching C and pointers to pointers. They will change the way you think about things like list management. The most important thing to remember is a pointer to pointer does not pointer to some "node"; it points to a pointer that points to a node. That is a pretty heady statement, but do some homework and it will make sense.

Answer 2

Edited:Use line below (or strncpy ) strcpy(n->tekst,t); instead of n->tekst[20] = t[20];

Initialize *test to null

struct Line lin, *tekst=NULL;

Use %s to print text in your program.

printf("%s", n->tekst);