Thread: Sorting by passing struct as argument

Hi,

I am writing a program that takes input fr stdin, breaks it into tokens of words, and then count the occurrence of the each word by sorting them into a tree of struct.

however when i tried to put the words into the struct, it kept giving root a NULL.

Here's my code:

Code:

void sortwords(char **warr, int wc, struct tnode *r){
	int c=0;
	for(c=0;c<wc;c++){
		sort(r, warr[c]);
	}
}


void sort(struct tnode *r, char *w){    
    if(r == NULL){
        r = (struct tnode *) malloc(sizeof(struct tnode));
        r->word = w;
        r->count = 1;
        r->left = NULL;
        r->right = NULL;
    }else{
    
        int cmp = strcmp(w, r->word);
        if(cmp<0){
            sort(r->left, w);
        }else if(cmp>0){
            sort(r->right, w);
        }else r->count++;
        
    }
}

and here's my struct and how i initialized it:

Code:

struct tnode {    char *word;
    int count;
    struct tnode *left;
    struct tnode *right;
};

int main (int argc, const char * argv[]) {
        ...
    struct tnode *root = NULL;
    ...
        //warray is an array of words, checked, no problem
        //wordsread is the number of words in the array
    sortwords(warray, wordsread, root);
    ...
    return 0;
}

Code:

void sort(struct tnode *r, char *w){    
    if(r == NULL){
        r = (struct tnode *) malloc(sizeof(struct tnode));

"r" is local to the function, thus when you change its value the caller doesn't notice it. That's why in sortwords() "r" will always point to NULL.
You have to pass a pointer to the object when you want that the caller also sees changes to the object. In your case your function prototype should be

Code:

void sort(struct tnode **r, char *w);

and you have to dereference accordingly.
The same is true for sortwords(). main() won't notice any changes to "root" inside sortwords().

Bye, Andreas