Calculating when trains will meet, iteratively and using OOP












1












$begingroup$


I've done a little bit of programming in the past, mostly just dabbling.
After a long time of not touching an IDE, I am getting back into it, with Visual Studio. I just threw this quick project together to make sure I remember how to use classes and objects before I start playing with bigger projects.



I would love a quick critique to make sure there arent any glaring poor practices that could develop into bad habbits in the future.



I know the standard way to solve this problem is just time = distance/(velA+velB), but what would the point of using objects be if i didnt have the objects do something and change their status in someway?



What do you think, looks good?



/*

The goal for this practice program is to solve the common math word problem below using classes and objects.
I feel that using iterations rather than the basic math formula to solve, although clearly less efficient, would be more true to thinking in terms of objects.

Train A, traveling X miles per hour (mph), leaves Westford heading toward Eastford, 260 miles away.
At the same time Train B, traveling Y mph, leaves Eastford heading toward Westford.
When do the two trains meet? How far from each city do they meet?

*/

#include "pch.h"
#include <iostream>
using namespace std;

class Train
{
public:
Train(int, int);
int getLocation();
void update();
private:
int location, velocity;
};

Train::Train(int loc, int vel) //to initialize the train object and set its location and velocity
{
location = loc;
velocity = vel;
}

int Train::getLocation() // returns the location of the train object
{
return location;
}

void Train::update() // updates the train object for one iteration
{
location += velocity;
}

int main()
{
int velA, velB, distance; //to take the values from the user input
int time; // to keep track of the number of iterations
// time is declared here so it can be used outside of the for loop


//input
cout << "Enter velocity of the train from Westford:n";
cin >> velA;
cout << "nnEnter velocity of the train from Eastford:n";
cin >> velB;
cout << "nnEnter the distance between Westford and Eastford:n";
cin >> distance;

//initialize each train
Train trainA(0, velA);
Train trainB(distance, 0 - velB); //location of trainB is distance because the distance between an x coordinate at 0 and another x coordinate is equal to the second x coordinate
//the velocity of trainB is the negative of velB because it is traveling in the opposite direction of trainA

//run the sim
for (time = 0; trainA.getLocation() < trainB.getLocation(); time++)
{
trainA.update();
trainB.update();
}

//output
cout << "nnThe Trains pass eachother after " << time << " hours."
<< "nAt that time, the Westford train is " << trainA.getLocation() << " miles from Westfordn"
<< "and the Eastford train is " << distance - trainB.getLocation() << " miles from Eastford.nn";

return 0;
}









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New contributor




Mickey Brenneman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • $begingroup$
    I guess a do while loop could make more sense here than a for loop?
    $endgroup$
    – Mickey Brenneman
    23 mins ago


















1












$begingroup$


I've done a little bit of programming in the past, mostly just dabbling.
After a long time of not touching an IDE, I am getting back into it, with Visual Studio. I just threw this quick project together to make sure I remember how to use classes and objects before I start playing with bigger projects.



I would love a quick critique to make sure there arent any glaring poor practices that could develop into bad habbits in the future.



I know the standard way to solve this problem is just time = distance/(velA+velB), but what would the point of using objects be if i didnt have the objects do something and change their status in someway?



What do you think, looks good?



/*

The goal for this practice program is to solve the common math word problem below using classes and objects.
I feel that using iterations rather than the basic math formula to solve, although clearly less efficient, would be more true to thinking in terms of objects.

Train A, traveling X miles per hour (mph), leaves Westford heading toward Eastford, 260 miles away.
At the same time Train B, traveling Y mph, leaves Eastford heading toward Westford.
When do the two trains meet? How far from each city do they meet?

*/

#include "pch.h"
#include <iostream>
using namespace std;

class Train
{
public:
Train(int, int);
int getLocation();
void update();
private:
int location, velocity;
};

Train::Train(int loc, int vel) //to initialize the train object and set its location and velocity
{
location = loc;
velocity = vel;
}

int Train::getLocation() // returns the location of the train object
{
return location;
}

void Train::update() // updates the train object for one iteration
{
location += velocity;
}

int main()
{
int velA, velB, distance; //to take the values from the user input
int time; // to keep track of the number of iterations
// time is declared here so it can be used outside of the for loop


//input
cout << "Enter velocity of the train from Westford:n";
cin >> velA;
cout << "nnEnter velocity of the train from Eastford:n";
cin >> velB;
cout << "nnEnter the distance between Westford and Eastford:n";
cin >> distance;

//initialize each train
Train trainA(0, velA);
Train trainB(distance, 0 - velB); //location of trainB is distance because the distance between an x coordinate at 0 and another x coordinate is equal to the second x coordinate
//the velocity of trainB is the negative of velB because it is traveling in the opposite direction of trainA

//run the sim
for (time = 0; trainA.getLocation() < trainB.getLocation(); time++)
{
trainA.update();
trainB.update();
}

//output
cout << "nnThe Trains pass eachother after " << time << " hours."
<< "nAt that time, the Westford train is " << trainA.getLocation() << " miles from Westfordn"
<< "and the Eastford train is " << distance - trainB.getLocation() << " miles from Eastford.nn";

return 0;
}









share|improve this question









New contributor




Mickey Brenneman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    I guess a do while loop could make more sense here than a for loop?
    $endgroup$
    – Mickey Brenneman
    23 mins ago
















1












1








1


1



$begingroup$


I've done a little bit of programming in the past, mostly just dabbling.
After a long time of not touching an IDE, I am getting back into it, with Visual Studio. I just threw this quick project together to make sure I remember how to use classes and objects before I start playing with bigger projects.



I would love a quick critique to make sure there arent any glaring poor practices that could develop into bad habbits in the future.



I know the standard way to solve this problem is just time = distance/(velA+velB), but what would the point of using objects be if i didnt have the objects do something and change their status in someway?



What do you think, looks good?



/*

The goal for this practice program is to solve the common math word problem below using classes and objects.
I feel that using iterations rather than the basic math formula to solve, although clearly less efficient, would be more true to thinking in terms of objects.

Train A, traveling X miles per hour (mph), leaves Westford heading toward Eastford, 260 miles away.
At the same time Train B, traveling Y mph, leaves Eastford heading toward Westford.
When do the two trains meet? How far from each city do they meet?

*/

#include "pch.h"
#include <iostream>
using namespace std;

class Train
{
public:
Train(int, int);
int getLocation();
void update();
private:
int location, velocity;
};

Train::Train(int loc, int vel) //to initialize the train object and set its location and velocity
{
location = loc;
velocity = vel;
}

int Train::getLocation() // returns the location of the train object
{
return location;
}

void Train::update() // updates the train object for one iteration
{
location += velocity;
}

int main()
{
int velA, velB, distance; //to take the values from the user input
int time; // to keep track of the number of iterations
// time is declared here so it can be used outside of the for loop


//input
cout << "Enter velocity of the train from Westford:n";
cin >> velA;
cout << "nnEnter velocity of the train from Eastford:n";
cin >> velB;
cout << "nnEnter the distance between Westford and Eastford:n";
cin >> distance;

//initialize each train
Train trainA(0, velA);
Train trainB(distance, 0 - velB); //location of trainB is distance because the distance between an x coordinate at 0 and another x coordinate is equal to the second x coordinate
//the velocity of trainB is the negative of velB because it is traveling in the opposite direction of trainA

//run the sim
for (time = 0; trainA.getLocation() < trainB.getLocation(); time++)
{
trainA.update();
trainB.update();
}

//output
cout << "nnThe Trains pass eachother after " << time << " hours."
<< "nAt that time, the Westford train is " << trainA.getLocation() << " miles from Westfordn"
<< "and the Eastford train is " << distance - trainB.getLocation() << " miles from Eastford.nn";

return 0;
}









share|improve this question









New contributor




Mickey Brenneman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I've done a little bit of programming in the past, mostly just dabbling.
After a long time of not touching an IDE, I am getting back into it, with Visual Studio. I just threw this quick project together to make sure I remember how to use classes and objects before I start playing with bigger projects.



I would love a quick critique to make sure there arent any glaring poor practices that could develop into bad habbits in the future.



I know the standard way to solve this problem is just time = distance/(velA+velB), but what would the point of using objects be if i didnt have the objects do something and change their status in someway?



What do you think, looks good?



/*

The goal for this practice program is to solve the common math word problem below using classes and objects.
I feel that using iterations rather than the basic math formula to solve, although clearly less efficient, would be more true to thinking in terms of objects.

Train A, traveling X miles per hour (mph), leaves Westford heading toward Eastford, 260 miles away.
At the same time Train B, traveling Y mph, leaves Eastford heading toward Westford.
When do the two trains meet? How far from each city do they meet?

*/

#include "pch.h"
#include <iostream>
using namespace std;

class Train
{
public:
Train(int, int);
int getLocation();
void update();
private:
int location, velocity;
};

Train::Train(int loc, int vel) //to initialize the train object and set its location and velocity
{
location = loc;
velocity = vel;
}

int Train::getLocation() // returns the location of the train object
{
return location;
}

void Train::update() // updates the train object for one iteration
{
location += velocity;
}

int main()
{
int velA, velB, distance; //to take the values from the user input
int time; // to keep track of the number of iterations
// time is declared here so it can be used outside of the for loop


//input
cout << "Enter velocity of the train from Westford:n";
cin >> velA;
cout << "nnEnter velocity of the train from Eastford:n";
cin >> velB;
cout << "nnEnter the distance between Westford and Eastford:n";
cin >> distance;

//initialize each train
Train trainA(0, velA);
Train trainB(distance, 0 - velB); //location of trainB is distance because the distance between an x coordinate at 0 and another x coordinate is equal to the second x coordinate
//the velocity of trainB is the negative of velB because it is traveling in the opposite direction of trainA

//run the sim
for (time = 0; trainA.getLocation() < trainB.getLocation(); time++)
{
trainA.update();
trainB.update();
}

//output
cout << "nnThe Trains pass eachother after " << time << " hours."
<< "nAt that time, the Westford train is " << trainA.getLocation() << " miles from Westfordn"
<< "and the Eastford train is " << distance - trainB.getLocation() << " miles from Eastford.nn";

return 0;
}






c++ beginner c++17






share|improve this question









New contributor




Mickey Brenneman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Mickey Brenneman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




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edited 2 mins ago









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New contributor




Mickey Brenneman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 1 hour ago









Mickey BrennemanMickey Brenneman

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New contributor




Mickey Brenneman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Mickey Brenneman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Mickey Brenneman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    I guess a do while loop could make more sense here than a for loop?
    $endgroup$
    – Mickey Brenneman
    23 mins ago




















  • $begingroup$
    I guess a do while loop could make more sense here than a for loop?
    $endgroup$
    – Mickey Brenneman
    23 mins ago


















$begingroup$
I guess a do while loop could make more sense here than a for loop?
$endgroup$
– Mickey Brenneman
23 mins ago






$begingroup$
I guess a do while loop could make more sense here than a for loop?
$endgroup$
– Mickey Brenneman
23 mins ago












0






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