Thread: Passing an array to a linked list

We probably need to see a bit more of your code to say exactly what is wrong. I suspect, however, that you are storing the address of pattern_small, and then overwriting pattern_small with something else. Since all the "word" pointers are pointing to the same thing, you only get one lot of data in your list [and if it's a local variable, once that is out of scope it may contain any random rubbish] Is this what you are seeing?

To solve that particular problem, you would need to allocate memory for the string and copy the string into your new structure. Or if your strings are of reasonable size, make a fixed size array of (say) 25 chars as your "word", and copy into that [note that malloc will often use large areas per allocation, so allocating 6-10 bytes or allocating 25 bytes will be the same amount - in visual studio .Net, malloc's smallest allocation appears to be 64 bytes].

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Mats

Yes, but what does the word pointer actually point to? Is that THE same place in memory or different places in memory?

You could easily check that with:

Code:

  ... 
ktemp = ( struct kword * ) malloc ( sizeof ( struct kword ));
ktemp->word = test;
ktemp->next = khead;
if (khead != 0 && khead != ktemp) {
   if (khead->word == ktemp->word) {
      printf("Words point to the same thing - this is not valid\n");
   }
}
khead = ktemp;

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Mats

Yes, and your point is?

Whether it's a linked list, and array, a hash-table or some other construction, if you are storing a string as a char *, each value of char * needs to point to a specific place which contains the string you refer to.

To make a simple example (try this on your system)

Code:

char *arr[3];

int main()
{
    char str[10] = "Hello";
    for(i = 0; i < 3; i++)
       arr[i] = str;
    strcpy(arr[2], "World");
    for(i = 0; i < 3; i++)
       printf("arr[%d]=%s\n", i, arr[i]);
    return 0;
}

Now do this (changes in red):

Code:

char *arr[3];

char *copystr(char *s)
{
    char *p = malloc(strlen(s)+1);
    strcpy(p, s);
    return p;
}

int main()
{
    char str[10] = "Hello";
    for(i = 0; i < 3; i++)
       arr[i] = copystr(str);
    strcpy(arr[2], "World");
    for(i = 0; i < 3; i++)
       printf("arr[%d]=%s\n", i, arr[i]);
    return 0;
}

The same would happen if "str" was inserted into a linked list. I choose an array to make it simple to understand.

--
Mats

Ok, so a pointer is just a memory address stored in a variable. One piece of memory can only store ONE value. Since your pattern_small variable is still the same place in memory each time you insert it into the linked list, although it contains different things each time you insert it, you actually overwrite the previous value each time, and this means that your linked list always contains one thing all throughout the list (many pointers, but all to the same place).

Think of a pointer as the address of a house, and the memory content is who lives in the rooms. If you have an address of 10 Kensington Avenue, London, then some people live there [and in this computer version of houses, you can't have more than one person living in each room]. You can change who lives there, but ultimately, only a certain set of people live there. So your pointer contains the address of the house. The string is the people living in the house. Does this make sense?

--
Mats