searching cstring for operators - c++

Question

For an assignment we are given an equation in reverse polish notation. For this example I will use: 2 3 8 * + $ The $ is for denoting the end of the expression. Using a stack, we output the answer.
I have been using:
getline(cin, input, '&');
input.c_str();
to read in the equation and then turn it into a c_string so I can look at the individual elements in the input.
After, I need to check for a few things. If the element is a digit I need to push it onto the stack. If it is white space I need to skip over it.
if (isdigit(input[i]))
{
push();
i++;
}
else if (isspace(input[i]))
{
i++;
}
Now is what has me stuck. If I hit an operator (in this case *) I need to pop the top two elements off the stack and 'operate them' and push the result back to the stack. However, I don't know of anything that would allow me to recognize that they are operators. It is probably a silly question, but help would be greatly appreciated.

Answer 1

You could create a functions map, and match the operators
std::map<char, void (*)(char *)> operators;
operators['*'] = &mult;
operators['+'] = &add;
...
and in your conditionnal statement
if (isdigit(input[i]))
{
push();
i++;
}
else if (isspace(input[i]))
{
i++;
}
else if (operators.find(input[i]) != operators.end())
{
operators[input[i]](input);
}
This way, you will be able to easily add new operators to your calculator.

Answer 2

there are only about 4 or 5 operators. check them with something like:
if (input[i] == '*') {...}

Answer 3

Well, there's no "built-in" way. I would just write an IsOperator(char ch) function. And then do something like:
int IsOperator(char ch)
{
// either add more operators or use a table, etc. here
if (ch == '+' || ch == '-' || ch == '/' || ch == '*')
return 1;
return 0;
}
If you have multi-char operators, like '==', it gets a little more complicated because you have to peek ahead, but the idea is the same.

Answer 4

If you are writing reverse polish notation calculator then on stack you have only numbers, or digits like in your example. BTW don't you accept multi-digits numbers?
Consider two cases, in both you need only std::stack<int> numbers;:
You have digit character c - just put number on stack:
code:
numbers,push(c - '0');
You have operator, let say '+', you replaces top two numbers by their sum:
code:
if (numbers.size() < 2) {
throw std::runtime_error("Too little numbers for +");
}
int a = numbers.top();
numbers.pop();
numbers.top() += a;
You have '$', check if there is only one number on stack, it is the result:
code:
if (numbers.size() != 1) {
throw std::runtime_error("There should be only one!");
}
int result = numbers.top();
numbers.pop();