A, B, or A and B (C++)

[32bites]

A am learning C++. I am currently trying to figure out a way for a user to have an option to do both A and B in addition to A or B.

So far i have put in an option to to A and B but I dont know the best way for it to actually do both.

// I will foo your bar

#include <iostream>
#include <cmath>
using namespace std;

int main ()
{
    float x1, x2, y1, y2, x, y, d;
    float midordist;

    cout << "nDo you want to find the midpoint (1), the distance (2), or both (3): ";
    cin >> midordist;

    if(midordist < 1 || midordist > 3) {
        cout << "nThe option you have entered is not valid";
        return 0;
    }

    cout << "nnPlese enter in the form (x1, y1) and (x2, y2)n";

    cout << "x1 = ";
    cin >> x1;

    cout << "y1 = ";
    cin >> y1;

    cout << "nThe first set is (" << x1 << ", " << y1 << ")nn";

    cout << "x2 = ";
    cin >> x2;

    cout << "y2 = ";
    cin >> y2;

    cout << "nThe second set is (" << x2 << ", " << y2 << ")nn";

    if(midordist == 1){ //find midpoint
        x = (x1 + x2) / 2;
        y = (y1 + y2) / 2;
        cout << "The midpoint is (" << x << ", " << y << ")";
    }

    if(midordist == 2){ //find distance
        x = pow(x2 - x1, 2);
        y = pow(y2 - y1, 2);
        d = sqrt(x+y);
        cout << "The distance between (" << x1 << ", " << y1 << ") and (" << x2 << ", " << y2 << ") is " << d;
    }

    return 0;
}

[snakey]

try printing both at the end instead of doing both at once like:

find midpoint

then

find distance 

then

print either a,b or ab

[32bites]

Thanks you your help I found a friend of mien to help me with this he came up with this:

// I will foo your bar

#include <iostream>
#include <cmath>
#include <cstdlib>
using namespace std;

int main ()
{
    float x1, x2, y1, y2, x, y, d;
    float midordist;

    cout << "nDo you want to find the midpoint (1), the distance (2), or both (3): ";
    cin >> midordist;

    if(midordist < 1 || midordist > 3) {
        cout << "nThe option you have enterd is not valid";
        return 0;
    }

    cout << "nnPlese enter in the form (x1, y1) and (x2, y2)n";

    cout << "x1 = ";
    cin >> x1;

    cout << "y1 = ";
    cin >> y1;

    cout << "nThe first set is (" << x1 << ", " << y1 << ")nn";

    cout << "x2 = ";
    cin >> x2;

    cout << "y2 = ";
    cin >> y2;

    cout << "nThe second set is (" << x2 << ", " << y2 << ")nn";

    if(midordist == 1 || midordist == 3){ //find midpoint
        x = (x1 + x2) / 2;
        y = (y1 + y2) / 2;
        cout << "The midpoint of (" << x1 << ", " << y1 << ") and (" << x2 << ", " << y2 << ") is (" << x << ", " << y << ")";
    }
    if(midordist == 3){
        cout << "n";
    }
    if(midordist == 2 || midordist == 3){ //find distance
        x = pow(x2 - x1, 2);
        y = pow(y2 - y1, 2);
        d = sqrt(x+y);
        cout << "The distance between (" << x1 << ", " << y1 << ") and (" << x2 << ", " << y2 << ") is " << d;
    }

    return 0;
}

This mainly ads || midordist == 3 to each if statement. along with adding n between the two results.