Question

我正在寻找一个可以合并到我正在开发的项目中的C ++类。我需要的功能是将字符串操作评估为数字形式：例如“2 + 3 * 7”应评估为23。

我确实意识到我所要求的是一种翻译，并且有建立它们的工具，我在CS的背景非常差，所以如果你能指出我已经准备好的课，我将不胜感激。 / p>

Answer 1

这应该完全符合你的要求。您可以在http://www.wowpanda.net/calc

进行实时测试

运营商优先级（5 + 5 * 5 = 30不是50）
Parens（（5 + 5）* 5 = 50）
以下运算符：+， - ，*，/

编辑：你可能想要删除底部的Abs（）;对于我的需要0 - 5应该是5而不是-5！

static bool Rpn(const string expression, vector<string> &output)
{
    output.clear();
    char *end;
    vector<string> operator_stack;
    bool expecting_operator = false;

    for (const char *ptr = expression.c_str(); *ptr; ++ptr) {
        if (IsSpace(*ptr))
            continue;

        /* Is it a number? */
        if (!expecting_operator) {
            double number = strtod(ptr, &end);
            if (end != ptr) {
                /* Okay, it's a number */
                output.push_back(boost::lexical_cast<string>(number));
                ptr = end - 1;
                expecting_operator = true;
                continue;
            }
        }

        if (*ptr == '(') {
            operator_stack.push_back("(");
            expecting_operator = false;
            continue;
        }

        if (*ptr == ')') {
            while (operator_stack.size() && operator_stack.back() != "(") {
                output.push_back(operator_stack.back());
                operator_stack.pop_back();
            }

            if (!operator_stack.size())
                return false; /* Mismatched parenthesis */

            expecting_operator = true;
            operator_stack.pop_back(); /* Pop '(' */
            continue;
        }

        if (*ptr == '+' || *ptr == '-') {
            while (operator_stack.size() && IsMathOperator(operator_stack.back())) {
                output.push_back(operator_stack.back());
                operator_stack.pop_back();
            }

            operator_stack.push_back(boost::lexical_cast<string>(*ptr));
            expecting_operator = false;
            continue;
        }

        if (*ptr == '*' || *ptr == '/') {
            while (operator_stack.size() && (operator_stack.back() == "*" || operator_stack.back() == "/")) {
                output.push_back(operator_stack.back());
                operator_stack.pop_back();
            }

            operator_stack.push_back(boost::lexical_cast<string>(*ptr));
            expecting_operator = false;
            continue;
        }

        /* Error */
        return false;
    }

    while (operator_stack.size()) {
        if (!IsMathOperator(operator_stack.back()))
            return false;

        output.push_back(operator_stack.back());
        operator_stack.pop_back();
    }

    return true;
} // Rpn

/***************************************************************************************/

bool Calc(const string expression, double &output)
{
    vector<string> rpn;

    if (!Rpn(expression, rpn))
        return false;

    vector<double> tmp;
    for (size_t i = 0; i < rpn.size(); ++i) {
        if (IsMathOperator(rpn[i])) {
            if (tmp.size() < 2)
                return false;
            double two = tmp.back();
            tmp.pop_back();
            double one = tmp.back();
            tmp.pop_back();
            double result;

            switch (rpn[i][0]) {
                case '*':
                    result = one * two;
                    break;

                case '/':
                    result = one / two;
                    break;

                case '+':
                    result = one + two;
                    break;

                case '-':
                    result = one - two;
                    break;

                default:
                    return false;
            }

            tmp.push_back(result);
            continue;
        }

        tmp.push_back(atof(rpn[i].c_str()));
        continue;
    }

    if (tmp.size() != 1)
        return false;

    output = Abs(tmp.back());
    return true;
} // Calc

/***************************************************************************************/

Answer 2

boost :: spirit附带一个计算器示例，可以满足您的需求： http://www.boost.org/doc/libs/1_33_1/libs/spirit/example/fundamental/ast_calc.cpp

Answer 3

muParser是用C ++编写的，可以满足您的需求。

Answer 4

C ++ in Action，除了是一本关于C ++的好书外，还包括一个完全可用的计算器，可以满足您的需求（实际上还有更多）。这本书可以在线免费获取

C ++简单操作（+， - ，/，*）评估类

4 个答案: