I have this assignment to make an infix calculator I think I have all the logical things figured out, but I am having trouble converting values. the return is a double, but the read in is a string and outputting as a double, I know that there is the difference is 48 from ascii value but I can't find where to take off this difference, I have tried a few different things and when I do 1+1 I either get 43 or -5 as my outputs. gonna keep banging my head but any help would be greatly appreciated! thank you! here is 2 of the 3 functions that assign my two stacks, 1 for digits and 1 for operators. anyone see something I don't?

``````double InFixCalculator::evaluate(string infix)
{
stack<char> opStack;
stack<double> valStack;
double test = 0;

for (unsigned int x = 0 ; x < infix.length() ; x++)
{
switch(infix[x])
{
case '1' :
case '2' :
case '3' :
case '4' :
case '5' :
case '6' :
case '7' :
case '8' :
case '9' :
valStack.push(infix[x]);
break;

case '(' :
opStack.push(infix[x]);
break;

/*case ')':
// Append/pop operators from stack until an ( is on top
while(opStack.top() != '(')
{
postFix += opStack.top();
opStack.pop();
}//end while
opStack.pop(); // Remove the open (
break;*/

case '+':
case'-':
case '*':
case '/':
if(opStack.empty())
opStack.push(infix[x]);

else if (precedence(infix[x]) > precedence(opStack.top()))
{
opStack.push(infix[x]);
}

else
{
while(!opStack.empty() && precedence(infix[x] <= precedence(opStack.top())))
{    execute(opStack,valStack);
opStack.push(infix[x]); // Push new operator onto stack.

}//end while
}//end if

break;

case ')':
while(opStack.top() != '(')
{
execute(opStack, valStack);
opStack.pop();
break;
}//end while

}//end switch
}//end for

while (!opStack.empty())
{
test += opStack.top();
opStack.pop();
}//end while
return test  ;
}//end evaluate

void InFixCalculator::execute(stack<char> &opStack, stack<double> &valStack)
{ //I used if statements instead of a switch case, i find them easier to follow
// and it prevents fanning of coding, please dont take any points off =)

const int ASCII_DIGIT_OFFSET = 48;
double operand2, operand1, result ;
char op; // declaring variables for operations decisions

operand2 = ((int)valStack.top() - ASCII_DIGIT_OFFSET);
valStack.pop();

operand1 = ((int)valStack.top()- ASCII_DIGIT_OFFSET);
valStack.pop();
//setting operands for the top 2 values on the Valstack
op = opStack.top();
opStack.pop();
//seting the type of operations for operand1 and operand 2 from opStack
if(op == '+')
{
result = operand1 + operand2 ;
}
if(op == '*')
{
result = operand1 * operand2 ;
}

if(op == '/')
{
if(operand2 == '0')
{
cout <<"Division by zero error!";//catching division by zero for error handling
}
else
{
result = operand1 / operand2 ;
}
}

if(op == '+')
{
result = operand1 + operand2 ;
}
if(op == '-')
{
result = operand1 - operand2 ;
}
valStack.push(result) ;// putting the result of top 2 operands onto the stack

}//end execute
``````

1) stack<double> valStack; declares valStack to be a stack that holds objects of type double, so why do you need a offset?

2) Numerical offsets can be cute/convenient to convert an individual digit to an int value but, unless you are working with indivual single digit numbers with values 0-9, it's a useless trick.

3) Your input is undoubtedly a string of characters of the form: 12 + 35. You will want to breakdown (aka parse) the input into substrings (frequently called tokens when talking about the process of parsing). At this point each token is still string. A token (string) can represent either an operand or an operator. Each operand token will need to be converted to a numeric type (int, double, float, whatever) to do mathematical calculations with them sooner or later. To convert a string of characters to a numeric type it's probably best if you use stringstreams, although function like atoi() or atol() or atof() are available (and may be easier to learn, but have drawbacks that make them less desireable to use).