c++四则运算
表达式计算
编写一个程序，可以计算算术表达式的值，对非法表达式要给出提示，并要求重新输入正确的表达式。（正确的表达式范例：123+45*6-78/9 或 (123+45)*(6-78)/9 等等）
题目要求：(1) 能正确计算出正确表达式的值。
(2) 要考虑表达式中的优先级。
相关知识：(1) 函数
(2) 结构



帮忙分析下。。
怎么用结构体？怎么构思。。

[解决办法]
结构应该知道是数据结构,堆栈之类的...
供参考

C/C++ code

#include <cstring>#include <stdexcept>#include <iostream>using namespace std;template<class T = char, size_t SIZE = 256>class Stack {public:    Stack (void) : m_uTop (0) {}    void Push (const T& tData) {        if (Full ())            throw overflow_error ("堆栈满！");        m_tData[m_uTop++] = tData;    }    T Pop (void) {        if (Empty ())            throw underflow_error ("堆栈空！");        return m_tData[--m_uTop];    }    const T& Top (void) const {        if (Empty ())            throw underflow_error ("堆栈空！");        return m_tData[m_uTop-1];    }    bool Full (void) const {        return m_uTop >= SIZE;    }    bool Empty (void) const {        return m_uTop == 0;    }private:    T m_tData[SIZE];    size_t m_uTop;};bool Prior (char cOp1, char cOp2) {    return (cOp1 == '*' || cOp1 == '/') && (cOp2 == '+' || cOp2 == '-');}// 将中缀表达式转换为后缀（逆波兰）表达式string Infix2Suffix (const string& strInfix) {    string strSuffix;    string::size_type uLength = strInfix.length ();    Stack<> stack;    for (string::size_type i=0; i<uLength; i++) {        // 数字和小数点        if (isdigit(strInfix[i])||strInfix[i]=='.')            // 直接入表达式            strSuffix += strInfix[i];        // 左括号        else if (strInfix[i] == '(')            // 直接入栈            stack.Push (strInfix[i]);        // 右括号        else if (strInfix[i] == ')') {            // 将相应左括号后的运算符栈入表达式            while (stack.Top () != '(')                (strSuffix += ' ') += stack.Pop ();            stack.Pop (); // 最后弹出左括号        }        // 运算符        else if (strchr ("+-*/", strInfix[i])) {            strSuffix += ' ';            // 该运算符不优先于栈顶，栈入表达式            while (! stack.Empty () && stack.Top () != '(' && ! Prior (strInfix[i], stack.Top ()))                (strSuffix += stack.Pop ()) += ' ';            // 优者入栈            stack.Push (strInfix[i]);        }    }    // 余者栈入表达式    while (! stack.Empty ())        (strSuffix += ' ') += stack.Pop ();    return strSuffix;}// 从后缀表达式中提取运算数double ExtractOperand (const string& strSuffix, string::size_type& uPos) {    double fOperand = 0;    // 处理整数部分    for (; isdigit (strSuffix[uPos]); uPos++)        fOperand = fOperand * 10 + strSuffix[uPos] - '0';    // 处理小数部分    if (strSuffix[uPos] == '.') {        string::size_type uDecimals = 0;        for (uPos++; isdigit (strSuffix[uPos]); uPos++) {            fOperand = fOperand * 10 + strSuffix[uPos] - '0';            uDecimals++;        }        for (; uDecimals; uDecimals--)            fOperand /= 10;    }    return fOperand;}double CalSuffix (const string& strSuffix) {    string::size_type uLength=strSuffix.length ();    Stack<double> stack;    for (string::size_type i = 0; i<uLength;i++) {        if (isdigit (strSuffix[i]))            stack.Push(ExtractOperand(strSuffix,i));        else if (strchr (" \t\r\n", strSuffix[i]))            i++;        else if (strchr ("+-*/", strSuffix[i])) {            double fOperand2 = stack.Pop ();            double fOperand1 = stack.Pop ();            switch (strSuffix[i]) {                case '+':                    stack.Push(fOperand1+fOperand2);                    break;                case '-':                    stack.Push(fOperand1-fOperand2);                    break;                case '*':                    stack.Push(fOperand1*fOperand2);                    break;                case '/':                    stack.Push(fOperand1/fOperand2);                    break;            }            i++;        }    }    return stack.Top ();}double CalInfix (const string& strInfix) {    return CalSuffix (Infix2Suffix (strInfix));}int main (void) {    cout << Infix2Suffix ("2+3*4") << endl;    cout << Infix2Suffix ("(2+3)*4") << endl;    cout << CalSuffix (Infix2Suffix ("2+3*4")) << endl;    cout << CalSuffix (Infix2Suffix ("(2+3)*4")) << endl;    cout << CalInfix ("(3-6.2)*12.34/((28.88+12.3-9.2))*17.21-(13.2+3.14)*2.97") << endl;    cout << (3-6.2)*12.34/((28.88+12.3-9.2))*17.21-(13.2+3.14)*2.97 << endl;    return 0;}