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;
}









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
    28 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
    28 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




share|improve this question








edited 8 mins ago









200_success

130k17154419




130k17154419






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

61




61




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
    28 mins ago




















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


















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






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












0






active

oldest

votes











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
});
});
}, "mathjax-editing");

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "196"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});






Mickey Brenneman is a new contributor. Be nice, and check out our Code of Conduct.










draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f215968%2fcalculating-when-trains-will-meet-iteratively-and-using-oop%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























0






active

oldest

votes








0






active

oldest

votes









active

oldest

votes






active

oldest

votes








Mickey Brenneman is a new contributor. Be nice, and check out our Code of Conduct.










draft saved

draft discarded


















Mickey Brenneman is a new contributor. Be nice, and check out our Code of Conduct.













Mickey Brenneman is a new contributor. Be nice, and check out our Code of Conduct.












Mickey Brenneman is a new contributor. Be nice, and check out our Code of Conduct.
















Thanks for contributing an answer to Code Review Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f215968%2fcalculating-when-trains-will-meet-iteratively-and-using-oop%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Сан-Квентин

Алькесар

Josef Freinademetz