c语言计算器实现，请问该怎么写

Question

要求能够处理括号，能够处理2*((1+2)+3+5/2)/2－6这种复杂公式。c语言实现，多谢!

Answer 1

代码如下，如有帮助，请采纳一下，谢谢。

#include <iostream>
#include <string>
#include <list>
using namespace std;
//判断是否合法
bool isValid(string &str)
{
    char spec[] = {'+','-','*','/','{','}','[',']','(',')'};
    //删除空格
    unsigned int i=0,j=0;
    char buf[100] = {0};
    while(i<str.length())
    {
        if(str.at(i)== ' ')
        {
            i++;
            continue;
        }
        else
        {
            buf[j++] = str.at(i);
            i++;
        }
    }
    str = buf;
 
    unsigned int len = str.length();
    if (str.at(len -1) != '=')
    {
        cout << "公式需以=结束"<< endl;
        return false;
    }else
        str = str.substr(0,len-1);
    //判断是否有非数字的内容
    i = 0;
    len = str.length();
    while(i < len)
    {
        if(str.at(i)< '0' || str.at(i) > '9'   )
        {
            for (j =0;j<10;j++)
            {
                if(str.at(i) == spec[j])
                    break;
            }
            if(j == 10)
            {
                cout << "公式中存在非法字符:" << str.at(i)<< endl;
                return false;
            }
            
        }
        i++;
    }
 
 
    //遇到3(2+3)、（2+3）3这种直接提示错误，不默认插入*
    i= 0;
    while(i < str.length()-1)
    {
        if(str.at(i) >= '0' && str.at(i) <= '9')
        {
            if(str.at(i+1)== '(' || str.at(i+1) == '[' || str.at(i+1) == '{')
            {
                cout << "在" << str.at(i) << "和" << str.at(i+1) << "之间需要添加运算符" << endl;
                return false;
            }
            
        }
        if(str.at(i)== ')' || str.at(i) == ']' || str.at(i) == '}')
        {
            if(str.at(i+1) >= '0' && str.at(i+1) <= '9')
            {
                str.insert(i+1,1,'*');
                cout << "在" << str.at(i) << "和" << str.at(i+1) << "之间需要添加运算符" << endl;
                return false;
            }
        }
        i++;
    }
    return true;
}
//判断是否是数字
bool isNmb(string s)
{
    unsigned int i = 0;
    if(s.empty())
        return false;
    if(s.at(0) == '-')
    {
        i=1;
        while(i<s.length())
        {
            if(s.at(i) == '.')
            {
                i++;
                continue;
            }
            if(s.at(i)< '0' || s.at(i) >'9')
                return false;
            i++;
        }
    }else
    {
        i = 0;
        while(i<s.length())
        {
            if(s.at(i) == '.')
            {
                i++;
                continue;
            }
            if(s.at(i)< '0' || s.at(i) >'9')
                return false;
            i++;
        }
    }
    return true;
}
 
//判断数字是否是整数
bool isZs(string s)
{
    unsigned int i = 0;
    while(i<s.length())
    {
        if(s.at(i) == '.')
            return false;
        i++;
    }
    return true;
}
 
//处理一个最简单的计算公式--TO BE TEST
bool DealSmple(string str,string &out)
{
    unsigned int i =0,j=0;
    
    //将字符串分隔
    list<string> vp;
    int prepos =0;
    while(i<str.length())
    {
        if (str.at(i) == '+'  || str.at(i)=='*' || str.at(i) == '/' )
        {
            string tmp = str.substr(prepos,i-prepos);
            vp.push_back(tmp);
            vp.push_back(str.substr(i,1));
            i++;
            prepos = i;
        }else if( str.at(i) =='-') //查找连续--的情况
        {
            if( (i == 0) || (vp.size() > 0 && (vp.back().compare("+") == 0 || vp.back().compare("-") == 0 || vp.back().compare("*") == 0 || vp.back().compare("/") == 0)) ) //如果上一字符是-号，那么把当前的-留给下面的字符
                i++;
            else
            {
                string tmp = str.substr(prepos,i-prepos);
                vp.push_back(tmp);
                vp.push_back(str.substr(i,1));
                i++;
                prepos = i;
            }
        }
        else
            i++;
    }
    if(prepos == str.length())
    {
        cout << "公式中存在以"<< str.at(str.length()-1) << "结尾的非法公式段"<<endl;
        return false;
    }
    vp.push_back(str.substr(prepos,str.length() - prepos));
 
    //如果只有一个元素，直接退出
    if(vp.size() == 1)
    {
        out = vp.front();
        return true;
    }
 
    //计算乘法除法
    list<string>::iterator it = vp.begin();
    list<string>::iterator tmp;
    for (;it != vp.end();)
    {
        if(it->compare("*")==0 )//先处理乘法和除法
        {
            if (it == vp.begin())
            {
                cout << "*出现在公式的开头"<< endl;
                return false;
            }
            it++;
            if (it == vp.end())
            {
                cout << "*出现在公式的末尾" << endl;
                return false;
            }
            string nmb2 = *it;
            tmp = it;
            tmp--;
            vp.erase(it);
            it = tmp;
            tmp--;
            vp.erase(it);
            it = tmp;
            string nmb1 = *it;
 
            //判断是否是数字
            if ((!isNmb(nmb1)) || (!isNmb(nmb2)))
            {
                cout << "存在多个连续的运算符或者数字个数不正确" << endl;
                return false;
            }
 
 
            char val[20]={0};
            //判断两个数是否都是整数
            if (isZs(nmb1) && isZs(nmb2))
            {
                int a = atoi(nmb1.c_str());
                int b = atoi(nmb2.c_str());
                int c = a*b;
                itoa(c,val,10);
            }else
            {
                double a = atof(nmb1.c_str());
                double b = atof(nmb2.c_str());
                double c = a*b;
                gcvt(c,10,val);
            }
 
            it = vp.insert(it,val);
            it++;
            vp.erase(it);
 
 
            it = vp.begin();
        }else if (it->compare("/") == 0)
        {
            if (it == vp.begin())
            {
                cout << "/出现在公式的开头"<< endl;
                return false;
            }
            it++;
            if (it == vp.end())
            {
                cout << "/出现在公式的末尾" << endl;
                return false;
            }
            string nmb2 = *it;
            tmp = it;
            tmp--;
            vp.erase(it);
            it = tmp;
            tmp--;
            vp.erase(it);
            it = tmp;
            string nmb1 = *it;
 
            //判断是否是数字
            if ((!isNmb(nmb1)) || (!isNmb(nmb2)))
            {
                cout << "存在多个连续的运算符或者数字个数不正确" << endl;
                return false;
            }
 
            char val[20]={0};
            //判断两个数是否都是整数
            if (isZs(nmb1) && isZs(nmb2))
            {
                int a = atoi(nmb1.c_str());
                int b = atoi(nmb2.c_str());
                if(b==0)
                {
                    cout << "除数为0"<< endl;
                    return false;
                }
                int c = a/b;
                itoa(c,val,10);
            }else
            {
                double a = atof(nmb1.c_str());
                double b = atof(nmb2.c_str());
                if(b==0.0)
                {
                    cout << "除数为0"<< endl;
                    return false;
                }
                double c = a/b;
                gcvt(c,10,val);
            }
 
            it = vp.insert(it,val);
            it++;
            vp.erase(it);
 
            it = vp.begin();
        }else
            it++;
    }
 
    //计算加减
    bool iszhs = true;
    for (it=vp.begin();it != vp.end(); it++)
    {
        if (!isZs(*it))
        {
            iszhs = false;
            break;
        }
    }
    if (vp.size() == 1)
    {
        out = vp.front();
        return true;
    }else
    {
        if (iszhs)
        {
            int sum = 0;
            it = vp.begin();
            string tt = *it;
            sum += atoi(tt.c_str());
            it++;
            while(it != vp.end())
            {
                string ts = *it;
                it++;
                if(it == vp.end())
                {
                    cout << "公式运算中出现以运算符结尾的情况"<< endl;
                    return false;
                }
                string nmb3 = *it;
                if (ts.compare("+") == 0)
                {
                    sum += atoi(nmb3.c_str());
                }else if (ts.compare("-") == 0)
                {
                    sum -= atoi(nmb3.c_str());
                }else
                {
                    cout << "公式出现不匹配的运算符" << endl;
                    return false;
                }
                it++;
            }
 
            char bufout[20] ={0};
            itoa(sum,bufout,10);
            out = bufout;
            return true;
 
        }else
        {
            double sum = 0.0;
            it = vp.begin();
            string tt = *it;
            sum += atof(tt.c_str());
            it++;
            while(it != vp.end())
            {
                string ts = *it;
                it++;
                if(it == vp.end())
                {
                    cout << "公式运算中出现以运算符结尾的情况"<< endl;
                    return false;
                }
                string nmb3 = *it;
                if (ts.compare("+") == 0)
                {
                    sum += atof(nmb3.c_str());
                }else if (ts.compare("-") == 0)
                {
                    sum -= atof(nmb3.c_str());
                }else
                {
                    cout << "公式出现不匹配的运算符" << endl;
                    return false;
                }
                it++;
            }
 
            char bufout[20] ={0};
            gcvt(sum,10,bufout);
            out = bufout;
            return true;
        }
 
    }
 
 
 
    return true;
}
 
//处理括号
bool DealKuhao(string &strSrc, char ch,char ch2)
{
    //先处理小括号
    while(1)
    {
        int posEnd = strSrc.find(ch);
        if(posEnd < 0)
            return true;
        else if(posEnd == 0)
        {
            cout << ch <<"缺少对应的左括号" << endl;
            return false;
        }else
        {
            //找到与之对应的大括号
            int i = posEnd -1;
            while(i >=0 && strSrc.at(i) != ch2)
                i--;
            if(i < 0)
            {
                cout << ch << "没有与之配对的左括号" << endl;
                return false;
            }
            string tmp = strSrc.substr(i+1,posEnd - i-1); //截取()之间的字符串
            string strout="";
            if(DealSmple(tmp,strout))  //处理()之间的公式，并返回处理结果
            {
                strSrc = strSrc.substr(0,i) + strout + strSrc.substr(posEnd+1,strSrc.length() - posEnd);
                //cout << "DealSmple " << ch << "  " << strSrc << endl;
            }else
                return false;
        }
    }
    return true;
}
 
 
 
 
int main()
{
    string strSrc,strout = "";
    cout << "请输入公式，以=号结束:" << endl;;
    getline(cin,strSrc);
 
    if(!isValid(strSrc))
        return 0;
 
    if(!DealKuhao(strSrc,')','('))
        return 0;
    if(!DealKuhao(strSrc,']','['))
        return 0;
    if(!DealKuhao(strSrc,'}','{'))
        return 0;
    DealSmple(strSrc,strout);
    cout << strout << endl;
 
    return 0;
 
}