Evaluate a simple string mathematical expression [closed]

Answer 1

Here is my shortest attempt. It took about 40 minutes to type up, you can play with it on ideone (link).

The code is very straightforward, assuming that you have at least a cursory familiarity with the basic recursive descent parsing technique.

#include 
#include 
using namespace std;
bool eval_expr(const char **pe, int &lhs, bool inside = false);
// gets the next char after skipping optional whitespace
char skip_ws(const char **pe) {
    while (**pe == ' ') ++(*pe);
    return **pe;
}
// evaluates a parenthesized expression or a number
bool eval_prim(const char **pe, int &res) {
    char c = skip_ws(pe);
    if (c == '(') {
        ++(*pe);
        if (!eval_expr(pe, res, true)) return false;
        ++(*pe);
        return true;
    }
    if (isdigit(c)) {
        res = 0;
        while (isdigit(c)) {
            res = 10*res + c - '0';
            c = *(++(*pe));
        }
        return true;
    }
    return false;
}
// evaluates a chain of + - * / operations
bool eval_expr(const char **pe, int &lhs, bool inside) {
    if (!eval_prim(pe, lhs)) return false;
    char op;
    while ((op = skip_ws(pe)) && (op == '+' || op == '-' || op == '*' || op == '/')) {
        ++(*pe);
        int rhs;
        if (!eval_prim(pe, rhs)) return false;
        switch (op) {
            case '+': lhs += rhs; break;
            case '-': lhs -= rhs; break;
            case '*': lhs *= rhs; break;
            case '/': lhs /= rhs; break;
        }
    }
    return inside ? op == ')' : !op;
}
// wrapper API to hide an extra level of indirection
bool evaluate(const char *e, int &result) {
    return eval_expr(&e, result);
}