Parsing String of symbols to Expression

Given an expression as a string str consisting of numbers and basic arithmetic operators(+, -, *, /), the task is to solve the expression. Note that the numbers used in this program are single-digit numbers and parentheses are not allowed.

Examples:

Input: str = “3/3+4*6-9”
Output: 16
Since (3 / 3) = 1 and (4 * 6) = 24.
So the overall expression becomes (1 + 24 – 9) = 16



Input: str = “9*5-4*5+9”
Output: 16

Approach: A Stack class is created to store both numbers and operators (both as characters). The stack is a useful storage mechanism because, when parsing expressions, the last item stored needs to be accessed frequently; and a stack is a last-in-first-out (LIFO) container.

Besides the Stack class, a class called express(short for expression) is also created, representing an entire arithmetic expression. Member functions for this class allow the user to initialize an object with an expression in the form of a string, parse the expression, and return the resulting arithmetic value.

Here’s how an arithmetic expression is parsed.
A pointer is started at the left and is iterated to look at each character. It can be either a number(always a single-digit character between 0 and 9) or an operator (the characters +, -, *, and /).

If the character is a number, it is pushed onto the stack. The first operator encountered is also pushed into the stack. The trick is subsequent operators are handled. Note that the current operator can’t be executed because the number that follows it hasn’t been read yet. Finding an operator is merely the signal that we can execute the previous operator, which is stored on the stack. That is, if the sequence 2+3 is on the stack, we wait until we find another operator before carrying out the addition.

Thus, whenever the current character is an operator (except the first), the previous number (3 in the preceding example) and the previous operator (+) are popped off the stack, placing them in the variables lastval and lastop. Finally, the first number (2) is popped and the arithmetic operation is carried on the two numbers (obtaining 5).
However, when * and / which have higher precedence than + and – are encountered, the expression can’t be executed. In the expression 3+4/2, the + cant be executed until the division is performed. So, the 2 and the + are put back on the stack until the division is carried out.

On the other hand, if the current operator is a + or -, the previous operator can be executed. That is when the + is encountered in the expression 4-5+6, it’s all right to execute the -, and when the – is encountered in 6/2-3, it’s okay to do the division. Table 10.1 shows the four possibilities.

Previous Operator Current Operator Example Action
+ or – * or / 3+4/ Push previous operator and previous number (+, 4)
* or / * or / 9/3* Execute previous operator, push result (3)
+ or – + or – 6+3+ Execute previous operator, push result (9)
* or / + or – 8/2- Execute previous operator, push result (4)

Below is the implementation of the above approach:

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <cstring>
#include <iostream>
using namespace std;
  
// Length of expressions in characters
const int LEN = 80;
  
// Size of the stack
const int MAX = 40;
  
class Stack {
private:
    // Stack: array of characters
    char st[MAX];
  
    // Number at top of the stack
    int top;
  
public:
    Stack()
    {
        top = 0;
    }
  
    // Function to put a character in stack
    void push(char var)
    {
        st[++top] = var;
    }
  
    // Function to return a character off stack
    char pop()
    {
        return st[top--];
    }
  
    // Function to get the top of the stack
    int gettop()
    {
        return top;
    }
};
  
// Expression class
class Express {
private:
    // Stack for analysis
    Stack s;
  
    // Pointer to input string
    char* pStr;
  
    // Length of the input string
    int len;
  
public:
    Express(char* ptr)
    {
        pStr = ptr;
        len = strlen(pStr);
    }
  
    // Parse the input string
    void parse();
  
    // Evaluate the stack
    int solve();
};
  
void Express::parse()
{
  
    // Character from the input string
    char ch;
  
    // Last value
    char lastval;
  
    // Last operator
    char lastop;
  
    // For each input character
    for (int j = 0; j < len; j++) {
        ch = pStr[j];
  
        // If it's a digit then save
        // the numerical value
        if (ch >= '0' && ch <= '9')
            s.push(ch - '0');
  
        // If it's an operator
        else if (ch == '+' || ch == '-'
                 || ch == '*' || ch == '/') {
  
            // If it is the first operator
            // then put it in the stack
            if (s.gettop() == 1)
  
                s.push(ch);
  
            // Not the first operator
            else {
                lastval = s.pop();
                lastop = s.pop();
  
                // If it is either '*' or '/' and the
                // last operator was either '+' or '-'
                if ((ch == '*' || ch == '/')
                    && (lastop == '+' || lastop == '-')) {
  
                    // Restore the last two pops
                    s.push(lastop);
                    s.push(lastval);
                }
  
                // In all the other cases
                else {
  
                    // Perform the last operation
                    switch (lastop) {
  
                    // Push the result in the stack
                    case '+':
                        s.push(s.pop() + lastval);
                        break;
                    case '-':
                        s.push(s.pop() - lastval);
                        break;
                    case '*':
                        s.push(s.pop() * lastval);
                        break;
                    case '/':
                        s.push(s.pop() / lastval);
                        break;
                    default:
                        cout << "\nUnknown oper";
                        exit(1);
                    }
                }
                s.push(ch);
            }
        }
        else {
            cout << "\nUnknown input character";
            exit(1);
        }
    }
}
  
int Express::solve()
{
    // Remove the items from stack
    char lastval;
    while (s.gettop() > 1) {
        lastval = s.pop();
        switch (s.pop()) {
  
        // Perform operation, push answer
        case '+':
            s.push(s.pop() + lastval);
            break;
        case '-':
            s.push(s.pop() - lastval);
            break;
        case '*':
            s.push(s.pop() * lastval);
            break;
        case '/':
            s.push(s.pop() / lastval);
            break;
        default:
            cout << "\nUnknown operator";
            exit(1);
        }
    }
    return int(s.pop());
}
  
// Driver code
int main()
{
  
    char string[LEN] = "2+3*4/3-2";
  
    // Make expression
    Express* eptr = new Express(string);
  
    // Parse it
    eptr->parse();
  
    // Solve the expression
    cout << eptr->solve();
  
    return 0;
}

chevron_right


Output:

4


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.