Starter Programmer help please. I'm using RECURSION

[karl kho]

The point of my code is to evaluate a mathematical expression that's already formed in a Binary Search Tree. : ((6+5)-3)*(2/1)=?

I've read other forums and they usually use cout, cin and << >> but I'm still in the learning process so this is basically primitive coding.

Anyway, The recursion is already correct (I think). My problem is how to return char* or should I say an element which has 2 or more digits.

typedef struct cell{
	char elem[max];
	struct cell *left;
	struct cell *right;
}celltype, *node;
node A;

char* evaluate(node *A)      //passed as evaluation(&A);
{
	char *c=(char*)malloc(sizeof(char),max);
	char *d=(char*)malloc(sizeof(char),max);
	
	strcpy(c,(*A)->elem);
	d=0;
	
	if((*A)->left!=NULL)
		c=evaluate(&((*A)->left));

	if((*A)->right!=NULL)
		d=evaluate(&((*A)->right));

	if(strcmp((*A)->elem,”+”)==0)* c=*c + *d;
	if(strcmp((*A)->elem,”-“)==0) *c=*d - *c;
	if(strcmp((*A)->elem,”*”)==0) *c=*c + *d;
	if(strcmp((*A)->elem,”/”)==0) *c=*d / *c;
	if(strcmp((*A)->elem,”=”)==0)
	{
		(*A)->right=(node)malloc(sizeof(celltype));
		(*A)->right->left=NULL;
		(*A)->right->right=NULL;
		(*A)->right->elem=*c;
	}
	return c;
}

and the reason that I'm using char is because the operators are also elements in different nodes.

The code above is actually the result when I remembered that you can't return 2-digit char.
The code below is before I remembered that. It works when all results are 1 digit numbers.

typedef struct cell{
	char elem;
	struct cell *left;
	struct cell *right;
}celltype, *node;
node A;

char evaluate(node *A)      //passed as “evaluation(&A);” in main
{
	char c,d;
	c=(*A)->elem;
	d=0;
	
	if((*A)->left!=NULL)
		c=evaluate(&((*A)->left));

	if((*A)->right!=NULL)
		d=evaluate(&((*A)->right));

	if((*A)->elem==0) c=c+d;
	if((*A)->elem==0) c=d-c;
	if((*A)->elem==0) c=c+d;
	if((*A)->elem==0) c=d/c;
	if((*A)->elem==0)
	{
		(*A)->right=(node)malloc(sizeof(celltype));
		(*A)->right->left=NULL;
		(*A)->right->right=NULL;
		(*A)->right->elem=c;
	}
	return c;
}

[ingenioushax]

Pass them by reference or their addresses. In this way, you can do something like the following:

#include <iostream>
using namespace std;

void change_chars(char&, char&);

int main()
{
	char a = 'C';
	char b = 'E';

	cout << "char a: " << a << "\nchar b:" << b << endl;
	change_chars(a, b);
	cout << "--[ After change_chars() ]--" << endl;
	cout << "char a: " << a << "\nchar b: " << b << endl;
	cin.get();
	return 0;
}

void change_chars(char& f, char& g)
{
	f = 'A';
	g = 'B';
}