Starting over Homework 2 Problem 5 a third time

bool infix_to_postfix(string infix, string& postfix)
// Returns false if conversion fails
{
    stack<char> stack_for_nonletters;

    postfix = "";

    for (int string_place = 0; string_place < infix.size(); string_place++)
    {
        char char_in_string = infix[string_place];

        // If it's simply a letter, add to the postfix expression
        if (char_in_string >= 'a' && char_in_string <= 'z')
        {
            // If there was two letters in a row in the infix form, we will return false
            if (string_place > 0 && (infix[string_place - 1] >= 'a' && infix[string_place - 1] <= 'z'))
                return false;

            if (string_place > 0 && infix[string_place - 1] )

            postfix += char_in_string;
        }

        // If there is an open brack, push it on the stack
        else if (char_in_string == '(')
        {
            stack_for_nonletters.push(char_in_string);
        }

        // If it is a closed bracket, keep popping from the stack until a closing pair open bracket is found
        else if (char_in_string == ')')
        {
            while (stack_for_nonletters.top() != '(')
            {
                postfix += stack_for_nonletters.top();
                stack_for_nonletters.pop();
            }

            // Now pop the open bracket
            stack_for_nonletters.pop();
        }

        // Let's now take into account the operatators
        else
        {
            // If the stack isn't empty and the current character in the string is of lower priority than one about to be added
            // We will pop the stack until an operand of lower priority
            while (!stack_for_nonletters.empty() && priority(char_in_string) <= priority(stack_for_nonletters.top()))
            {
                postfix += stack_for_nonletters.top();
                stack_for_nonletters.pop();
            }

            stack_for_nonletters.push(char_in_string);
        }
    }

    // If we have finished going through the infix expression,
    while (!stack_for_nonletters.empty())
    {
        postfix += stack_for_nonletters.top();
        stack_for_nonletters.pop();
    }

    return true;
}