How can I compare the next character which the file contains? - c++

Question

I am getting few problems when reading code. This is the text file.
2X^6+3X^3+4X^0=0
5X^6+X^2+X^1-4X^0=0
I am getting a proper input for the first line but in second line first I need to ignore. I searched here and found how to use it and it's work to get to next line ignoring all the left over characters of first line.
You can see in second line with X there is no integer, now problem is the second while loop is running continuously. If I add 1 in text file with the X the file reads perfectly. Also how can I put a condition to satisfy this that when there is directly X or -X, it should store 1 or -1 and goes to next character? Also you can see ^ I have to store this in a variable whereas I should ignore it but didn't how to ignore it?
Thanks in advance
int main()
{
int in;
int power;
char x;
char f;
fstream fin;
fin.open("input1.txt");
list l1,l2;
while(fin.peek() != 61)
{
fin>>in;
fin>>x;
fin>>f;
fin>>power;
cout<<in<<endl<<x<<endl<<f<<endl<<power<<endl;
l1.addtoend(in,power,x);
cout<<endl;
}
fin.ignore(2,'\n');
while(fin.peek() != 61)
{
fin>>in;
fin>>x;
fin>>f;
fin>>power;
cout<<in<<endl<<x<<endl<<f<<endl<<power<<endl;
l2.addtoend(in,power,x);
cout<<endl;
}
l1.display();
l2.display();
}

Answer 1

Unfortunately, this is not so simple as expected.
We need to split up the task into smaller parts.
What you want to do, is splitting your equation in terms and extract from this the coefficients and exponents.
Splitting up something in similar parts is also called tokenizing.
So, your equation consists of terms, which all follow the same pattern. First an optional sign, followed by the coefficients, then a “X^”, and, at the end the exponent (which may or may not have a sign).
And since all terms have the same pattern, we can find them with a so-called regex. C++ supports this functionality. Also for splitting up a text in smaller tokens/terms/pattern-matches, we have a special iterator std::sregex_token_iterator. Like any other iterator in C++, it iterates over the source string and extracts (and copies) all matched patterns.
OK, then we found already a solution for the first sub task. Extract all terms and put them into a std::vector. We will use the std::vectors range constructor, to do this, while defining the variable.
The next step is to get the coefficient. Here we need some special handling, because the coefficient can be omitted with an assumed 1. Using this assumption, we will read the term and convert the coefficient to an integer. And because we want to do that in one statement, we use std::transform from the STL’s algorithm library.
Getting the exponents is easier. We simply convert anything in a term following the ‘^’-sign to an integer. We again use std::transform to work on all terms in one statement.
Last but not least, we will get the right-hand-side of the equation and convert it also to an integer.
Please note:
All this can be done also with float type values
We could also allow spaces in the equation
For that, we would simple modify the std::regex-string.
Please see the complete example below:
#include <iostream>
#include <string>
#include <vector>
#include <iterator>
#include <regex>
#include <algorithm>
#include <iomanip>
int main() {
std::string equation{ "5X^6+X^2+X^1-4X^0=0" };
const std::regex re(R"(([+-]?\d?X\^[+-]?\d+))");
std::vector<std::string> terms{ std::sregex_token_iterator(equation.begin(), equation.end(), re,1),std::sregex_token_iterator() };
std::vector<int> coefficients(terms.size());
std::vector<int> exponents(terms.size());
int rightHandSite{ 0 };
// Everything in front of X is the coefficient. Handle special case, when no digit is given
std::transform(terms.begin(), terms.end(), coefficients.begin(), [](const std::string& s) {
std::string temp = s.substr(0U, s.find('X'));
if (1 == temp.size() && !std::isdigit(temp[0])) temp += '1';
return std::stoi(temp); });
// Get all exponents
std::transform(terms.begin(), terms.end(), exponents.begin(), [](const std::string & s) {
return std::stoi(s.substr(s.find('^') + 1)); });
// Get right Hand site of equation
rightHandSite = std::stoi(equation.substr(equation.find('=') + 1));
// Show result
std::cout << "\nEquation: " << equation << "\n\nFound '" << terms.size() << "' terms.\n\nCoeffient Exponent\n";
for (size_t i = 0U; i < terms.size(); ++i)
std::cout << std::right << std::setw(9) << coefficients[i] << std::setw(10) << exponents[i] << "\n";
std::cout << "\n --> " << rightHandSite << "\n";
return 0;
}
There are many other possible solutions. But maybe it will give you some idea on what you could do.