class function call.

Question

3.00/5 (1 vote)

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hi,
I'm creating a binary search tree.

class code:

C++

class bstnode{

private:
	int key;
	bstnode *left;
	bstnode *right;

public:
	//bstnode
	bstnode()
		{key=-1; left=0; right=0;}
//set right
	void set_right(bstnode *r)
		{right = r;}

main code:

C++

//declare a bstnode
bstnode(node2);

//set right of node
	node.set_right(node2);

i get erreor below:

IntelliSense: no suitable conversion function from "bstnode" to "bstnode *" exists

how can i solve it?

Thank you,

Posted 5-Jun-13 10:30am

m.r.m.40

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Comments

nv3 5-Jun-13 16:54pm

The second line of your main code is either incomplete or not correct. Did you want to say:
bstnode node;
And in line 5 you probably wanted to write:
node.set_right (&node2);

m.r.m.40 5-Jun-13 16:59pm

when i do as you said,
when i print node.right, it gives me the address of node2 not the key of it.

1 solution

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CPallini · Accepted Answer · 2013-06-05T10:52:00

Solution 1

You should provide a pointer to (that is the address of) bstnode object as (the only) argument to set_right method (while you are passing the object itself), change from

Quote:
node.set_right(node2);

to

node.set_right(&node2);

Please note:

Quote:
//declare a bstnode
bstnode(node2);

is not a valid C++ variable declaration, change to

C++

bstnode node2;

Posted 5-Jun-13 10:52am

CPallini

Comments

m.r.m.40 5-Jun-13 16:55pm

i did it,
when i printed node.right
it gave me the address of node.right not the key of node2
how can i see the key of node2 ?
thank you,

CPallini 5-Jun-13 16:59pm

node.right->key

m.r.m.40 5-Jun-13 17:02pm

thank you CPallini,
i didn't know about "->" expression.
what is its name(i want to search and learn about it)

CPallini 5-Jun-13 17:16pm

It is the 'arrow operator'.

Sergey Alexandrovich Kryukov 5-Jun-13 20:07pm

Sure, a 5.
—SA