私は多項式の計算を行っていますが、コードを進めるときに助けが必要です。
今のところ、用語といくつかの関数を含むリンクリストとして表現したポリノムクラスのみを作成しました（今のところ、多項式関数のみを読み取り、出力します）。
これは、多項式を読み取って出力するだけのメインプログラムです。
#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;
}
そしてヘッダファイル 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
編集
JonH のおかげですべてのコンパイル エラーが修正されましたが、入力文字がリストに正しく挿入されていないため、読み取り関数が機能していません。皆さんにとって些細なことかもしれませんが、助けていただければ幸いです。
ありがとう！