Trouble filling a dynamic c array - c

Question

I am having issues with filling an array of strings that was created dynamically. The array basically contains two strings and I would like to print them both with their lengths. I am getting a very weird result whereby the memory locations are getting mixed up. Please see the code below.
Any suggestions would greatly be appreciated. Thanks.
void printWords(char **words, int numberOfWords)
{
int index;
for (index = 0; index < numberOfWords; index++)
{
printf("%s, %d\n", &(*words)[index], (int)strlen(&(*words)[index]));
}
}
void fillWords(char **words)
{
*words = malloc(2 * sizeof(char *));
char hello[] = {"Hello"};
(*words)[0] = (char)malloc(strlen(hello) * sizeof(char));
strcpy(&(*words)[0], hello); //Copy word into array
char world[] = {"Worldz"};
(*words)[1] = (char)malloc(strlen(world) * sizeof(char));
strcpy(&(*words)[1], world); //Copy word into array
}
int main (int argc, const char * argv[])
{
char *words;
fillWords(&words);
printWords(&words, 2);
return 0;
}
The expected output should be
Hello, 5
Worldz, 6
however I am getting
HWorldz, 7
Worldz, 6

Answer 1

I think you are getting confused between char * and char **.
Also, be sure to allocate enough memory for the null-termination character at the end of the string.
Here is my solution:
#include <stdio.h>
#include <string.h>
#include <malloc.h>
void printWords(char **words, int numberOfWords)
{
int index;
for (index = 0; index < numberOfWords; index++)
{
printf("%s, %d\n", words[index], (int)strlen(words[index]));
}
}
char ** createWords()
{
char ** words;
// Allocate memory for an array of pointers, length 2.
words = malloc(2 * sizeof(char *));
char hello[] = {"Hello"};
words[0] = malloc((strlen(hello)+1) * sizeof(char));
strcpy(words[0], hello); //Copy word into array
char world[] = {"Worldz"};
words[1] = malloc((strlen(world)+1) * sizeof(char));
strcpy(words[1], world); //Copy word into array
return words;
}
int main (int argc, const char * argv[])
{
char **words;
words = createWords();
printWords(words, 2);
return 0;
}
I renamed fillWords to createWords and made it return a pointer instead of taking a pointer as an argument. If you really want fillWords to take a pointer as an argument, you can do it, but the argument will have to be of type char ***.

Answer 2

First problem: you can't put 2 words in a char * (well ... you can, but not the way you're doing; and the way to do it is nothing like your code).
You need an "array of strings", or, more C-like, an array of char *:
char *words[2]; /* words[0] and words[1] are pointers to char */
char *wrong; /* wrong[0] is a char; wrong[1] is a char */
So, change the definition of words in your main and check/edit all the other functions for correctness.
int main (int argc, const char * argv[])
{
char *words[2];
fillWords(words);
printWords(words, 2);
return 0;
}

Answer 3

Here is the solution. I was getting confused between char* and char. What I want was an array of chars.
void printWords(char **words, int numberOfWords)
{
int index;
for (index = 0; index < numberOfWords; index++)
{
printf("%s, %d\n", words[index], (int)strlen(words[index]));
}
}
void fillWords(char **words)
{
*words = malloc(2 * sizeof(char *));
char hello[] = {"Hello"};
words[0] = malloc(strlen(hello) * sizeof(char));
strcpy(words[0], hello); //Copy word into array
char world[] = {"Worldz"};
words[1] = malloc(strlen(world) * sizeof(char));
strcpy(words[1], world); //Copy word into array
}
int main (int argc, const char * argv[])
{
char *words;
fillWords(&words);
printWords(&words, 2);
return 0;
}