Data Loss when trying to copy char* in C - c

Question

I have been working on a project in C and I am having trouble when trying to copy char* using strcpy/memcpy/strncpy, none of these seem to work. The problem that is arising is that the words that are around 8 or more characters long are not being copied completely.
typedef struct wordFrequency {
char * word;
int frequency;
struct wordFrequency *left, *right;
} *node;
node setnode(char * word) {
node newNode = (node)malloc(sizeof(node));
newNode->word = (char*)malloc(sizeof(word));
strcpy(newNode->word, word); //This is where I'm having trouble
newNode->frequency = 1;
newNode->right = NULL;
return newNode;
}
The code above is what I believe is the main cause for error, but I don't know where to fix it. I have tried messing with the sizes, but that didn't work.
If possible can someone explain to me a way to copy all characters or if I did not allocate enough space?

Answer 1

This program is an mcve that shows how to properly allocate and initialize each node in your linked list:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define ARRAY_SIZE(array) \
(sizeof(array) / sizeof(array[0]))
typedef struct wordFrequency {
char *word;
int frequency;
struct wordFrequency *left, *right;
} node;
node *setnode(char *word) {
node *newNode = malloc(sizeof(node));
newNode->word = malloc(strlen(word) + 1);
strcpy(newNode->word, word);
newNode->frequency = 1;
newNode->right = NULL;
return newNode;
}
int main() {
char *wordList[] = {"one", "two", "three"};
node nodeHead;
node *nodePrev = &nodeHead;
node *nodeNext;
for (int index = 0; index < ARRAY_SIZE(wordList); index++) {
nodeNext = setnode(wordList[index]);
nodePrev->right = nodeNext;
nodeNext->left = nodePrev;
nodePrev = nodeNext;
}
for (node *nodePtr = nodeHead.right; nodePtr != NULL; nodePtr = nodePtr->right) {
printf("word = %s, frequency = %d\n", nodePtr->word, nodePtr->frequency);
}
return 0;
}
Output
word = one, frequency = 1
word = two, frequency = 1
word = three, frequency = 1
Note
This program has no error checking and does not free the allocated memory. This code should not be used in a production environment.
Replies to Questions in Comments
I replaced *node with node in the typedef because that allows me to declare instances of node. The other syntax only allows pointers to node.
I use an instance of node instead of node * for nodeHead because any attempt to change its address will be an error.
I use nodePrev to traverse the list and also to provide a target for left in the returned nodes. I initialize nodePrev to &nodeHead because it is the start of the list. I set nodePrev to nodeNext because that's how I chose to traverse the list during initialization. I could have used
nodePrev = nodePrev->right;
and achieved the same effect.
I only implemented list handling so that I could create a self-contained example that would run without changes. You can safely ignore it.
If you want to see good linked list code, I recommend the linux kernel implementation.