finding a pair with given sum in BST

Question

struct node
{
    int val;
    struct node *left, *right;
};

// Stack type 

struct Stack
{
    int size;
    int top;
    struct node* *array;
};

struct Stack* createStack(int size)
{
    struct Stack* stack =
        (struct Stack*) malloc(sizeof(struct Stack));
    stack->size = size;
    stack->top = -1;
    stack->array =
        (struct node**) malloc(stack->size * sizeof(struct node*));
    return stack;
}

What does this statement do?

stack->array =
    (struct node**) malloc(stack->size * sizeof(struct node*));

What will be the memory representation of it?

[Please see this discussion on why not to cast the return value of `malloc()` and family in `C`.](http://stackoverflow.com/q/605845/2173917). — Sourav Ghosh, Jul 12 '16 at 13:30
http://linux.die.net/man/3/malloc let us know what is unclear in it. — Sourav Ghosh, Jul 12 '16 at 13:31

score 0 · Answer 1 · answered Jul 12 '16 at 13:31

0

stack->array =
 (struct node**) malloc(stack->size * sizeof(struct node*));

struct node** returns a pointer to a pointer (to the stack)

stack->size is the amount of items of the stack sizeof(struct node*) is the size of a pointer to a node.

So it creates an array of pointers, where each pointer points to one element within the stack.

answered Jul 12 '16 at 13:31

Michel Keijzers

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so i guess this wiil be like :- – rohit singh Jul 13 '16 at 12:15

score 0 · Answer 2 · answered Jul 12 '16 at 13:32

0

The above statement allocates space for an array of struct node *, i.e. pointers to struct node.

Each pointer in this array can then point to an instance of struct node.

answered Jul 12 '16 at 13:32

dbush

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finding a pair with given sum in BST

2 Answers2