Why does the root change its data?

Question

Here is are the three structures that I'm using and for instance when my program gets 'the' as the first word, it makes *rt->str = the. However when the next word is read, the key is equal to to the *rt->str and I don't understand why. I'm a c programmer beginnner and this has really stoppped me in my tracks.

struct node {
    char *str;
    int occ;
    struct node *sibling;
    struct node *child;
};

struct node* root;

struct node* getNew(char word[100]) {
    struct node *newNode;
    newNode = (struct node *)malloc(sizeof(struct node));
    newNode->str = word;
    newNode->sibling = NULL;
    newNode->child = NULL;
    newNode->occ = 0;
    return newNode;
}

struct node* insert( char key[100], struct node **rt ){

    if(*rt == NULL) {
        *rt = getNew(key);
        printf("This is the key in the root: %s\n", (*rt)->str);
        return *rt;
    }else{
        printf("root word: %s\n", (*rt)->str);
        exit(0);
    }

    struct node *leaf = *rt;
    int n = 0;
    int i;
    char w2[100];
    strcpy(w2, key);

    printf("root word: %s\n", (*rt)->str);

    for(i = 0; i < strlen((leaf)->str); i++) {
        printf("%c %c \n", (leaf)->str[i], key[i]);
        if((key[0] == (leaf)->str[i])) {
            n++;
            key = key + 1;
            printf("key is: %s \n", key);
        }
    }

    if(key[0] == 0) {
        printf("key is empty \n");
    }

    printf("This is the word after for loop: %s \n", key);
    exit(0);
}

`if (...) return ..; else exit();` makes the rest of the function unreachable. — mch, Mar 01 '17 at 13:01
Learn to use a debugger. You won't get far by asking on SO every single time your code doesn't do what you wanted. — rustyx, Mar 01 '17 at 13:02
Also in `getNew`, `newNode->str = word;` You probably want to use [`strdup`](http://en.cppreference.com/w/c/experimental/dynamic/strdup) here. — 001, Mar 01 '17 at 13:05

score 1 · Answer 1 · answered Mar 01 '17 at 13:04

This:

newNode->str = word;

doesn't copy the string (as in, the characters that build up the string), it just copies the location of a string, which is an argument. That location will not remain valid when the function exits, so this gives you undefined behavior when you access it later.

C does not support assigning arrays, and arrays are not pointers.

Why does the root change its data?

1 Answers1