C语言：实现一个包含文本文件中字符串（单词）的BST的数组

Question

I am trying to create a small program that reads words from a text file of an unknown size and stores those words into an array of Binary Search Trees ( BST ) . Each index within the array represents the length of the words within that BST tree.

For example, index 0 contains no words but index 1 contains a BST tree with words that are one letter long and index 5 contains a BST tree with words that are 5 letters long etc. All BST trees are balanced by comparing two strings to determine if the new string is greater or smaller than the root string and then assigned accordingly.

My original code contains opaque objects ( void pointers ). But, I have included a smaller version of the program that I am trying to understand. I included printf statements to show my debugging approach because the program keeps crashing. I have been working on this for hours on end daily and can't get it to run for the life of me. For some reason I couldn't determine if I was using the pointers properly so after about 5 different rewrites of this code, I decided to go with just the basics, but I can't seem to get this to work either.

Please help, this is draining me. Thank you for your generosity and consideration in helping me with this in advance.

My output is as follows:

A CHECKPOINT
B CHECKPOINT
C CHECKPOINT
1 CHECKPOINT
2 CHECKPOINT

The code is as follows:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct my_string{
char* data;
struct my_string *left, *right;
} My_string;

void init( My_string* Root, char* data );

int main(int argc, char* argv[]){
    My_string* myStringArray[ 30 ] = {NULL};
    /*My_string* Root = NULL;*/
    FILE *fp = NULL;
    char new_string[ 30 ];
    fp = fopen( "dictionary.txt", "r");
    int string_length = 0;
    printf( "A CHECKPOINT\n");
    while( fscanf( fp, "%1024s" , new_string ) == 1 ){
        printf( "B CHECKPOINT\n");
        string_length = strlen( new_string );
        printf( "C CHECKPOINT\n");
        init( myStringArray[ string_length ], new_string );
        printf( "D CHECKPOINT\n");
    }
    printf( "" );
    fclose(fp);
    return 0;
}

void init( My_string* Root, char* data ){
    printf( "1 CHECKPOINT\n");
    int compare = 0;
    if( Root == NULL ){
        printf( "2 CHECKPOINT\n");
        (*Root).data = ( My_string* )malloc( sizeof( My_string ));
         printf( "3 CHECKPOINT\n");
        if( !Root ) exit(1);
        Root->data = data;
        Root->left = Root->right = NULL;
    }
    else{
        if( compare = strncmp( data, Root->data, 36 ) == 0 )return;
        else if( compare == -1 ) init( Root->left, data );
        else init( Root->right, data );
    }
}

Thanks again!

Answer 1

Two general advices:

1. Instead of debug-output you can use debugger to find the exact location of the error (learn how to use gdb , for example). You can use debug-output, of course, but it can take more time and you have to clean up after this.
2. Don't ignore compiler warnings.

My compiler says:

a.c:37:22: warning: assignment from incompatible pointer type [-Wincompatible-pointer-types]
(*Root).data = ( My_string* )malloc( sizeof( My_string ));

Here you try to dereference Root and assign a value to data field. Since Root is NULL , the program crashes here. It looks like you intended to assign value to Root here, but made a type of kind. So it should be something like this:

Root = ( My_string* )malloc( sizeof( My_string ));

BTW, you have another problem in the code: when you pass Root as function parameter, it won't be changed after you exit the function:

My_string* Root = NULL;
init(Root, data);
// Root is NULL here

One way to fix this is to pass pointer to Root :

init(&Root, data);
void init( My_string** Root_ptr, char* data ){
    ...
}

and modify the code accordingly.

Another way is to change init signature and make it return a newly-created Root . I don't understand the scenario where you need to init an existing tree, so it seems a natural thing to do.