searching cstring for operators

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.

just off-hand, is your input guaranteed to be seperated by whitespace? — WhozCraig, Oct 03 '12 at 18:43
@WhozCraig Yessir, it will always have a space between each value. — ViscousRandom, Oct 03 '12 at 20:03
What if you input contains multi digits numbers: `12 321 + $` ? — PiotrNycz, Oct 03 '12 at 20:18
@PiotrNycz I still need to think about how to account for that as well. — ViscousRandom, Oct 03 '12 at 20:36
@ViscousRandom just do not put digit at once on stack - but keep counting the number - unless you get any non digit character - then put the number on stack, BTW for reverse polish notation you need only stack of numbers - nothing more - see my response. — PiotrNycz, Oct 03 '12 at 20:54

score 1 · Answer 1 · answered Oct 03 '12 at 18:22

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.

score 0 · Accepted Answer · answered Oct 03 '12 at 18:16

0

there are only about 4 or 5 operators. check them with something like:

if (input[i] == '*') {...}

answered Oct 03 '12 at 18:16

none

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pfft. i'd use a switch and strtok(), but odds are runtime-libraries are rather limited for their assignment. Then again, i see isdigit(), isspace(), etc. – WhozCraig Oct 04 '12 at 03:46

score 0 · Answer 3 · answered Oct 03 '12 at 18:19

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.

score 0 · Answer 4 · answered Oct 03 '12 at 20:51

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();

searching cstring for operators

4 Answers4