I have been given a lab assignment that should take a prefix expression such as *+-23/427 and use recursion to solve its infix equivilant ((2-3)+(4/2))*7. So far, the code I have will work for an equation with 2 or 3 numbers. Any more and it starts returning weird results. Its probably a simple mistake on my part, but I'm not too great with recursion and I'm having a hard time figuring out whats going wrong. Any help would be greatly appreciated.

```
#include <math.h>
#include <iostream.h>
#include <string.h>
class Evaluate {
/* does prefix evaluations of expressions holding single digit values & formatted correctly
infix and postfix will be added next year
*/
public:
// each of the following returns an integer representing the result from
// evaluating the expression exp using the indicated type of notation
int preFix(char exp[]); // prefix notation
private:
// recursively implements preFix
int doPreFix (char exp[], int &left);
// other private methods
int doOp (int num1, char op, int num2); // returns num1 op num2
}; // Evaluate
void main()
{
Evaluate it;
char expres1[] = "+27";
cout <<"01234567890123456789\n" <<expres1 <<endl;
cout <<endl <<expres1 <<" = " <<it.preFix (expres1)<<endl<<endl;//outputs 9
cout << "--------------------------------------------------------" << endl;
char expres2[]= "+-235";
cout <<"01234567890123456789\n" <<expres2 <<endl;
cout <<endl <<expres2 <<" = " <<it.preFix (expres2)<<endl<<endl;//outputs 4
cout << "--------------------------------------------------------" << endl;
char expres3[]= "+1*2-41";
cout <<"01234567890123456789\n" <<expres3 <<endl;
cout <<endl <<expres3 <<" = " <<it.preFix (expres3)<<endl<<endl;//should output 7(outputs a 2)
cout << "--------------------------------------------------------" << endl;
char expres[]= "*+-23/427";
cout <<"01234567890123456789\n" <<expres <<endl;
cout <<endl <<expres <<" = " <<it.preFix (expres)<<endl<<endl;//should output 7 (outputs a 0)
}//main
int Evaluate::preFix(char exp[])
{
int left= 0;
return doPreFix (exp, left);
} // Evaluate::inFix
int Evaluate::doOp (int num1, char op, int num2)
{
switch (op) {//performs the correct operation accoding to the operator present
case '+': return num1 + num2;
case '-': return num1 - num2;
case '*': return num1 * num2;
case '/':
if (num2==0) {
cerr <<"Error: division by 0" <<endl;
return 0;
} // if
return num1 / num2;
default:
cerr <<"Error in formula: +-*/ expected, found a " <<op <<endl;
return 0;
} // switch
} // Evaluate::doOp
int Evaluate::doPreFix (char exp[], int &left)
{//assumes the first character is always an operator,
//and that each number is a positive integer only one digit long
int firstNum, secondNum, result;
char op;
if(exp[left] == '+' || exp[left] == '-' || exp[left] == '*' || exp[left] == '/')
{//takes in first operator
op = exp[left];//sets op to the position in the string
cout << " first op: " << op << "\t" << "left: " << left << endl;
left++;//increments left
firstNum = doPreFix(exp, left);
}
else if(exp[left] != NULL)//if the character isn't an operator, then it must be a number
{
firstNum = int (exp[left]) - int ('0');//convert from a character to an integer
cout << " firstNum: " << firstNum << "\t" << "left: " << left << endl;
left++;
return (firstNum);
}
if(exp[left] == '+' || exp[left] == '-' || exp[left] == '*' || exp[left] == '/')
{
op = exp[left];
cout << " second op: " << op << "\t"<< "left: " << left << endl;
left++;
secondNum = doPreFix(exp, left);
}
else if(exp[left] != NULL)//if the character isn't an operator, then it must be a number
{//and if the character isnt null.
secondNum = int (exp[left]) - int ('0');
cout << " second Num: " << secondNum << "\t" << "left: " << left<< endl;
left++;
doOp(firstNum, op, secondNum);
// return secondNum;
}
cout << "First Number: " << firstNum << " Operator: " << op << " Second number: "
<< secondNum;
result = doOp(firstNum, op, secondNum);
return result;
} // Evaluate::doPreFix
```