What is the difference between these codes involving a structure?

Question

I have a structure node which has 2 pointers small and large and an int val. I intend to make a binary tree out of it later, but before that I just tried to insert only one element in three different ways and then print it.

The 1st way works fine when I use a double ptr during insertion-

struct node
{
    struct node *small;
    struct node *large;
    int val;
};

void insert(node **root, int data)
{
    node *ptr=(node*)malloc(sizeof(node));
    
    ptr->val=data;
    ptr->large=NULL;
    ptr->small=NULL;
    
    if(*root==NULL)
    {
        *root=ptr;
        return;
    }
}

void print(node *ptr)
{
    cout<<ptr->val;
}


int main()
{
    int a[]{5};
    
    node *root=NULL;
    
    insert(&root,a[0]);
    print(root);
    
    return 0;
}

In 2nd way the root node is global and single ptr is used during insertion, it also works just fine-

struct node
{
    struct node *small;
    struct node *large;
    int val;
};

node *root=NULL;          //root node is global here

void insert(int data)
{
    node *ptr=(node*)malloc(sizeof(node));
    
    ptr->val=data;
    ptr->large=NULL;
    ptr->small=NULL;
    
    if(root==NULL)
    {
        root=ptr;
        return;
    }
}

void print()
{
    cout<<root->val;
}


int main()
{
    int a[]{5};

    node *root=NULL;
    
    insert(a[0]);
    print();
    
    return 0;
}

In the 3rd way root node is not global and a single ptr is used during insertion. But this time it gives some runtime error-

struct node
{
    struct node *small;
    struct node *large;
    int val;
};

void insert(node *root, int data)
{
    node *ptr=(node*)malloc(sizeof(node));
    
    ptr->val=data;
    ptr->large=NULL;
    ptr->small=NULL;
    
    if(root==NULL)
    {
        root=ptr;
        return;
    }
}

void print(node *root)
{
    cout<<root->val;
}


int main()
{
    int a[]{5};
    
    node *root=NULL;
    
    insert(root,a[0]);
    print(root);
    
    return 0;
}

I don't understand what the issue is with the 3rd approach and how is it different from the other two? I'm using the DevC++ compiler.

why you dropped the & in the 3rd method? Won't work without it: insert(&root... — Bing Wang, Mar 10 '21 at 21:15
You're passing `root` by value in **way 3** so it's still NULL when you call print. You can [pass by reference](https://stackoverflow.com/questions/10240161/reason-to-pass-a-pointer-by-reference-in-c) instead or stick to **way 1**. — Woodford, Mar 10 '21 at 21:42

score 1 · Accepted Answer · answered Mar 10 '21 at 22:05

Looks like you're still learning pointers? If so, consider code like this:

int a = 10;
int *b = &a;
int **c = &b;

You get:

c -> b -> a=10

If you pass a to a function and change it, just the value is passed and the original is unchanged:

void fa(int a) { a = 11; }
...
fa(a);
printf("%d\n", a);  // -> 10

If you pass b to a function, you can change a, but not b:

void fb(int *b) { *b = 11; b = NULL; }
...
fb(b);  // Also works with fb(&a)
printf("%d\n", a);   // -> 11
printf("%d\n", *b);  // -> 11

If you pass c, you can change a and b:

void fc(int **c) { **c = 11; *c = NULL; }
...
fc(c);  // Also works with fb(&b)
printf("%d\n", a);   // -> 11
printf("%d\n", *b);  // Segmentation fault dereferencing NULL

If you want a function to update a variable, in C++ better to declare it as a reference:

void fa(int& a) { a = 11; }
...
fa(a);
printf("%d\n", a);  // -> 11

Yeah I'm still learning. I was trying to modify the ptr itself without passing it as reference, that was the mistake I made. Thanks that cleared a lot of my doubts. — Jack Stevens, Mar 10 '21 at 22:14

What is the difference between these codes involving a structure?

1 Answers1