How do I use IF in FOR correctly ?

[Arquon]

Hello, I need to do some exercises, in which I need to use IF in FOR, and im really having trouble with it. The result I need to get in exercises :
1st exercise : if y1=y4, it has to say "Crosses", otherwise "***". In 2nd one, if y>0, it has to say "***".
What I have wrote in 1st and 2nd exercise, but no idea on how to use if to get the correct answers on y. Here is what I have wrote without if, and the results I need to get:

#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
int main()
{
 double x,y;
 for (x=15;x<=25;x++)
 {
    y=sqrt(x*x+3*x-500);
    cout << x << "  " << setprecision(3) << y << endl;
 }
    return 0;
}

What I need to get:
x y
15 ***
16 ***
17 ***
18 ***
19 ***
20 ***
21 2.00
22 7.07
23 9.90
24 12.17
25 14.14

2nd exercise:

#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
int main()
{
 double x,y1,y2;
 for (x=0;x<=10;x++)
 {
    y1=2*x+2;
    y2=3*x+1;
    cout << x << "  "  << y1 << "  " << y2 << endl;
 }
    return 0;
}

What i need to get :
x y1 y2
1 2 1 ***
1 4 4 Crosses
2 6 7 ***
3 8 10 ***
4 10 13 ***
5 12 16 ***
6 14 19 ***
7 16 22 ***
8 18 25 ***
9 20 28 ***
10 22 31 ***

Would be really grateful for some help here.

[epicmike]

So for your first problem you are getting an error for y. The reason that is happening is because you are trying to take the square root of a negative number. Try using an if statement to display *** if (x*x+3*x-500) < 0.

#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
int main()
{
	double x, y;
	cout << setw(3) << left << "x" << "y" << endl;
	for (x = 15; x <= 25; x++)
	{
		y = sqrt(x*x + 3 * x - 500);
		if ((x * x + 3 * x - 500) < 0)
		{
			cout << x << " " << "***" << endl;
		}
		else
		cout << fixed << setprecision(0) << x << " " << fixed << setprecision(2) << y << endl;
	}
	return 0;
}

[48T7M4Gy]

First one:

calculate z= x*x + 3*x etc
if ( z < 0)
   display ***
else
   display sqrt(z)

[epicmike]

Please feel free to ask if you don't understand some of this. I tried to use only the stuff in the beginning of a college c++ book.

#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
int main()
{
	double x, y1, y2;
	cout << setw(3) << left << "x" << setw(3) << left << "y1" << setw(3) << left << "y2" << endl;
	for (x = 0; x <= 10; x++)
	{
		y1 = 2 * x + 2;
		y2 = 3 * x + 1;
		if (y2 != y1)
		{
			cout << setw(3) << left << x << setw(3) << left << y1 
                                << setw(3) << left << y2 << "***" << endl;
		}
		else
			cout << setw(3) << left << x << setw(3) << left << y1
                               << setw(3) << left << y2 << "Crosses" << endl;
	}
	return 0;
}

[48T7M4Gy]

#include <iostream>
#include <iomanip>
#include <cmath>

using namespace std;
int main()
{
 double x,y, z;
 for (x=15;x<=25;x++)
 {
    z = x * x + 3 * x - 500;
    if (z < 0)
        cout << x << "   ***";
    else
        cout << x << "  " << setprecision(3) << sqrt(z);

     cout << endl;
 }
    return 0;
}

[Arquon]

Thank you for your help. I didnt think its that complicated with all those setw and left. And I have one question about 1st exercise. When I get y, why if(y<0) doesnt work ? I know its square root, but no idea why it doesnt work.

[48T7M4Gy]

Unless you are working in the complex plane, the square root of -ve numbers does not exist.

The inverse of the square root is a squared number which is always positive.
If a= sqrt(x) then x = a*a and if a is -ve or +ve x will always be +ve because +ve * +ve is +ve, and -ve * -ve is +ve.
So there is no way to get a -ve for x unless complex/imaginary i is allowed, which it isn't. :)

[Arquon]

Okay, thank you ^^