Why am I not able to display strings with the following code? - c

Question

If I put an "&" in the printf, it will at-least print the names of the countries and it will crash. Otherwise, it crashes immediately. My guess is, there is something wrong with the for loop. Let me know, if I am wrong.
//Implementation file//
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
struct list
{
char *item;
int length;
};
typedef struct list LIST; //aliasing the structure//
LIST *create() //constrcutor//
{
LIST *L = (LIST*)malloc(sizeof (LIST));
L->length=0;
return L;
}
void insert(LIST* L, char *newitem)
{
L->item=(char*)malloc(sizeof(char));
strcpy(L->item, newitem);
(L->length)++;
}
void display(LIST *L)
{
int i;
for(i=0;i<(L->length);i++)
{
printf("Elements are: %s\n", L->item[i]);
}
}
int main ()
{
LIST *L;
L=create();
insert(L, "America");
insert(L, "Brazil");
display(L);
free(L->item);
free(L);
return 0;
}

Answer 1

L->item=(char*)malloc(sizeof(char));
strcpy(L->item, newitem);
You are allocating only one byte, but you try to copy a whole string into L->item, resulting in undefined behaviour. This should be:
L->item=malloc(sizeof(char)*(strlen(newitem)+1));
strcpy(L->item, newitem);
You are also calling the function insert twice, which will allocate new memory for L->item and loose the pointer to the previous allocated space -> Memory leak. You should add list elements instead of overwriting the first one all the time, see #DavidKernin's answer.

Answer 2

There's a gigantic error in your code: when inserting new elements you're supposed to create a new list element and initialize it..
With the function
void insert(LIST* L, char *newitem)
{
L->item=(char*)malloc(sizeof(char));
strcpy(L->item, newitem);
(L->length)++;
}
called like this:
insert(L, "America");
insert(L, "Brazil");
you're always writing the same element for length and allocating new memory (without even using/storing a pointer to the old allocated one.. which is lost!).
You should rather do something like:
//Implementation file//
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
struct list
{
char *item;
struct list* next;
};
typedef struct list LIST; //aliasing the structure//
LIST *create() //constrcutor//
{
LIST *L = (LIST*)malloc(sizeof (LIST));
L->next = 0;
return L;
}
LIST* insert(LIST* L, char *newitem)
{
L->item=(char*)malloc(sizeof(char)*strlen(newitem));
strcpy(L->item, newitem);
LIST *newL = (LIST*)malloc(sizeof (LIST));
newL->next = 0;
L->next = newL;
return newL;
}
void display(LIST *L)
{
int i;
for(i=0; L->next != 0;L = L->next) // Assumes there's at least 1 item
{
printf("Elements are: %s\n", L->item);
}
}
int main ()
{
LIST *startElement;
startElement=create();
LIST *lastElement = insert(startElement, "America");
lastElement = insert(lastElement, "Brazil");
display(startElement);
// Free all memory.. exercise :)
return 0;
}
Try it out interactively: http://ideone.com/n8tRtv
Otherwise the purpose of linked lists would be entirely defeated.. who wants a list which loses the previous elements when you insert new ones?
Notice that the above code
Doesn't free memory
Always assumes there's at least an item in the list
If you want a complete code, well.. fix it. If you understand how it works it should be rather feasible to fix it entirely. Good luck!

Answer 3

The following modification of your code will allow you to insert strings and print them. Since that is what you asked for, that is what you shall get. If you had some sort of specific data structure in mind (perhaps a linked list etc.) when coding this up, please specify.
struct list
{
char **item;
int length;
};
struct list now holds an array of pointers to char *.
LIST *create()
{
LIST *L = (LIST*)malloc(sizeof (LIST));
L->item = (char **)malloc(sizeof(char *));
L->length=0;
return L;
}
This would be the proper way to malloc for your list.
void insert(LIST* L, char *newitem)
{
L->item[L->length] = (char *)malloc(sizeof(char *));
strcpy(L->item[L->length++], newitem);
}
To insert you malloc for a new char pointer, which is placed in the end of your array of char pointers. The inserted string is copied and length is incremented.
And don't forget to free()!

Answer 4

L->item=(char*)malloc(sizeof(char));
is one problem in the code

Answer 5

The specifier in the printf statement is incorrect. You are using %s which expects a char* but you are passing a char. As a result the printf implementation is treating the character value as an address and promptly running off into invalid memory and crashing
To fix this you need to switch to using %c
printf("Elements are: %c\n", L->item[i]);
Note that this will cause every char to be printed out on a different line. Most likely you want something like the following
printf("Elements are: %s\n", L->item[i]);
int i;
for(i=0;i<(L->length);i++)
{
printf("%c", L->item[i]);
}
printf("\n");
Note that if your intent is to have the item member be a null terminated char* then the simplest fix is to do the following
printf("Elements are: %s\n", L->item);