多项式计算器

Question

我正在做一个多项式计算器，随着代码的进展，我需要一些帮助。

现在我只制作了 polinom 类，我将其表示为带有术语和一些函数的链接列表（现在仅读取和打印多项式函数）。

这是现在仅读取多项式并打印它的主程序：

#include "polinom.h"

int main()

{

polinom P1;
bool varStatus = false;
char var = '\0', readStatus = '\0';

cout << "P1 = ";
P1.read(readStatus, var, varStatus); // i don't need readStatus yet as i haven't implemented the reset and quit functions 

cout << "\n\nP = ";
P1.print(var);

getch();
return 0;
}

头文件 polinom.h：

#ifndef _polinom_h
#define _polinom_h

#include <iostream>
#include <list>
#include <cstdlib>
#include <cctype>
#include <cstdio>
#include <conio.h>


using namespace std;

class polinom 
{
class term
{
    public:
        int coef;
        int pow;

        term() 
        {
            coef = 1;
            pow = 0;
        }    
};

list<term> poly;
list<term>::iterator i;

public:

    bool printable(char c) 
    {

        return (
                  ((int(c) > 42 && int(c) < 123) || isspace(c)) && int(c) != 44 && int(c) != 46 && int(c) != 47 && 
                  int(c) != 58 && int(c) != 59 && 
                  int(c) != 60 && int(c) != 61 && int(c) != 62 && int(c) != 63 && int(c) != 64 && int(c) != 65 && 
                  int(c) != 91 && int(c) != 92 && int(c) != 93 && int(c) != 95 && int(c) != 96
                ); 
    }


    void read(char &readStatus, char &var, bool &varStatus)
    {

        term t; // term variable to push it into the list of terms
        char c, lc, sign; // c = current char, lc = lastchar and sign the '+' or '-' sign before a coefficient
        int coef, pow; //variables to pass the coef and power to term t
        bool coefRead = false, powRead = false; //reading status of coef and power 

        while (c != '\r') { //we read characters until carriage return
            c = getch(); // get the new imputed char

            if (tolower(c) == 'r' || tolower(c) == 'q') { //if the user inputed r or q we reset the input or quit the program
                    readStatus = c; //pass current char value to readStatus so the program will know what to do next
                    return; //aborting the reading process
            }

            else 
            {
                if (printable(c)) cout << c; //print on screen only the correct characters

                if (!coefRead && !powRead) //we set term coef to the inputed value
                {                    
                    if (isdigit(c)) { 
                        if (isdigit(lc)) coef = coef * 10 + int(c); //if the last char was also a digit we multiply the last value of coef by 10 and add current char
                        else {                                    
                            if (sign == '-')  coef = -(int(c));//if the current coef has '-' before we set coef to it's negative value 
                            else              coef = int(c);   //this means a new term's coef is read
                    }
                    if (!isdigit(c) && isdigit(lc)) coefRead = true; //if the last char was a digit and we reached the var name we stop reading the coefficient
                }

                else if (coefRead && !powRead) //after coefficient is read we get the term's varname and power 
                {
                    if (isdigit(c)) { // just like in the case with coefficient we read the power until the current char is not a digit
                        if (isdigit(lc)) pow = pow * 10 + int(c);
                        else pow = int(c);
                    }

                    else if (isalpha(c) && isdigit(lc) && !varStatus) { //if the last char was a digit and the current not we reached the var name
                    var = c;                                            //also even though the variable is inputed more than once we save it only once
                    varStatus = true; //we mark the var name as read
                    }
                    else {
                        if (isdigit(lc)) powRead = true;
                    }   
                }

            else {
                if (c == '+' || c == '-') { // if a sign was inputed it means a new term is coming and we push the current term to the list and reset 
                    t.coef = coef;          // coefRead and powRead so we can read another term 
                    t.pow = pow;
                    poly.push_back(t);
                    sign = c;
                    coefRead = false;
                    powRead = false;
                }
            }

           lc = c; // we save the last character

            }
        } 
    }

    void print(char var)
    {
        for ( i=poly.begin() ; i != poly.end(); i++ ) { //going through the entire list to retrieve the terms and print them 

            if (i == poly.end() - 1) { // if we reached the last term 
                if (*(i->pow == 0) //if the last term's power is 0 we print only it's coefficient
                    cout << *(i->coef);
                else 
                    cout << *(i->coef) << var << "^" << *(i->pow); //otherwise we print both
            }

            else {
                if (*(i->coef > 0) //if the coef value is positive 
                    cout << *(i->coef) << var << "^" << *(i->pow) << " + "; //we also add the '+' sign
                else 
                    cout << *(i->coef) << var << "^" << *(i->pow) << " - "; // otherwise we add '-' sign
            }
        }
    }

};


#endif                    

编辑

感谢 JonH，所有编译错误现在已修复，但读取函数无法用作输入字符未正确插入到列表中。我知道这对你们来说可能微不足道，但如果你们能帮助我那就太好了。

谢谢！

Answer 1

我发现您的代码中缺少许多花括号和右括号。在花了几分钟修复了至少 10 个这些问题之后，我认为如果我帮助你学习钓鱼，而不是给你鱼作为今晚的晚餐，你会得到更好的服务。

你的代码写得就像意识流一样。当您构建代码时，您的思维会跳来跳去，思考您需要构建的其他内容以及您刚刚编写的内容引入的新需求。当你想到这些事情时，你就去写它们，然后回到原来的地方。在不知不觉中，您已经通过跳来跳去、到处写一些位来编写了数百行代码。这样做的问题是，您不可能在不遗漏一些语法位的情况下继续处理这样的代码段。

您应该采用更具迭代性的方法来编写代码。具体如何执行此操作取决于经验，但这里有一些指导：

1. 首先用一些（最好是 1 个）核心方法和成员变量删除类声明。
2. 编译。您会遇到链接器错误等，但您不应该遇到任何语法错误，例如缺少括号或分号。在继续之前修复您发现的任何问题。
3. 实现您刚刚存根的方法/函数。编译&修复非链接器错误。
4. 当您想到上述步骤中出现的次要或相关要求时，请在代码中编写注释，例如 // TODO: Implement bool DoTheThing(int); 但先不要实现它们。
5. 返回到步骤 1，使您正在处理的工作范围尽可能有限且基本。如果没有干净的编译，切勿超出编译步骤。

重复直到你完成了所有的事情。在此过程中您可能会编译 50 次或更多次。

Answer 2

你的根本问题是你写了一堆代码，没有一点一点地测试它，没有考虑它。当您作为初学者编写时，应该尝试一次添加一点并确保它可以编译。即使作为高级程序员，模块化也是设计和代码编写过程中极其重要的一部分。

也就是说，这里有一些关于您发布的代码的提示：

1. 您的函数 printable 非常丑陋，因此无法调试或理解。
2. 嵌套 if 语句的数量表明了设计缺陷。
3. 您的 if (isdigit(c)) 语句中缺少结束大括号。
4. 在同一行声明（尤其是初始化）多个变量是不好的形式。

Answer 3

这些编译错误肯定在错误消息中具有与其关联的行号。您是否尝试过查看指示的行以查看缺少的内容？如果这没有帮助，请发布编译器的完整错误输出，以便我们可以了解错误是什么。

Answer 4

您的读取函数中缺少一些花括号。

我在这里重做了：

 void read(char &readStatus, char &var, bool &varStatus) 
{ 

    term t; // term variable to push it into the list of terms 
    char c, lc, sign; // c = current char, lc = lastchar and sign the '+' or '-' sign before a coefficient 
    int coef, pow; //variables to pass the coef and power to term t 
    bool coefRead = false, powRead = false; //reading status of coef and power  

    while (c != '\r') { //we read characters until carriage return 
        c = getch(); // get the new imputed char 

        if (tolower(c) == 'r' || tolower(c) == 'q') 
        { //if the user inputed r or q we reset the input or quit the program 
                readStatus = c; //pass current char value to readStatus so the program will know what to do next 
                return; //aborting the reading process 
        } 

        else  
        { 
            if (printable(c)) 
               cout << c; //print on screen only the correct characters 

            if (!coefRead && !powRead) //we set term coef to the inputed value 
            {                     
                if (isdigit(c)) 
                   {  
                    if (isdigit(lc)) 
                       coef = coef * 10 + int(c); //if the last char was also a digit we multiply the last value of coef by 10 and add current char 
                    else 
                      {                                     
                        if (sign == '-')  
                           coef = -(int(c));//if the current coef has '-' before we set coef to it's negative value  
                        else              
                           coef = int(c);   //this means a new term's coef is read 
                      } //end else 
                    }//end if isdigit(c)
                if (!isdigit(c) && isdigit(lc)) 
                   coefRead = true; //if the last char was a digit and we reached the var name we stop reading the coefficient 
            }  //end if

            else if (coefRead && !powRead) //after coefficient is read we get the term's varname and power  
            { 
                if (isdigit(c)) 
                   { // just like in the case with coefficient we read the power until the current char is not a digit 
                    if (isdigit(lc)) 
                       pow = pow * 10 + int(c); 
                    else 
                         pow = int(c); 
                    } 

                else if (isalpha(c) && isdigit(lc) && !varStatus) 
                     { //if the last char was a digit and the current not we reached the var name 
                         var = c;                                            //also even though the variable is inputed more than once we save it only once 
                          varStatus = true; //we mark the var name as read 
                     } 
                else 
                     { 
                     if (isdigit(lc)) 
                        powRead = true; 
                     }    
            } //end else if 

        else 
             { 
             if (c == '+' || c == '-') 
                { // if a sign was inputed it means a new term is coming and we push the current term to the list and reset  
                  t.coef = coef;          // coefRead and powRead so we can read another term  
                  t.pow = pow; 
                  poly.push_back(t); 
                  sign = c; 
                  coefRead = false; 
                  powRead = false; 
                } 
              } 

       lc = c; // we save the last character 

        } //end else
    }  //end while
} //end function

编辑

我还修复了打印功能：

 void print(char var) 
    { 
        for ( i=poly.begin() ; i != poly.end(); i++ ) { //going through the entire list to retrieve the terms and print them  

            if (i == poly.end()) { // if we reached the last term  
                if (i->pow == 0) //if the last term's power is 0 we print only it's coefficient 
                    cout << i->coef;
                else  
                    cout << i->coef << var << "^" << i->pow; //otherwise we print both 
            } 

            else { 
                if (i->coef > 0) //if the coef value is positive  
                    cout << i->coef << var << "^" << i->pow << " + "; //we also add the '+' sign 
                else  
                    cout << i->coef << var << "^" << i->pow << " - "; // otherwise we add '-' sign 
            } 
        } 
    }