Equation returning 1.#QO - c

Question

I've tried searching for anything similar about my issue on several websites, including this one, but none I've found so far are similar. After searching about the term 1.#QO, I found something about quiet NaN, but I'm new to C in general, so I don't really understand the issue .
I'm trying to take the x and y values of a joystick, and when then use a formula for distance to find the distance between the joystick's position, and the joystick's natural resting position (0,0).
If it matters, I'm trying to do this in RobotC.
#pragma config(Hubs, S1, HTMotor, none, none, none)
#pragma config(Sensor, S1, , sensorI2CMuxController)
#pragma config(Motor, mtr_S1_C1_1, DriveMotor, tmotorTetrix, openLoop)
//*!!Code automatically generated by 'ROBOTC' configuration wizard !!*//
#include "JoystickDriver.c"
int calculateDrivePower(int joyStickX, int joyStickY)
{
if (joyStickX != 0 & joyStickY != 0)
{
int joyDistance = (sqrt(pow((joyStickX - 0),2)+ pow((-joyStickY - 0),2)));
joyDistance = ((joyDistance/127)*100);
return (joyDistance);
}
else
{
return 0;
}
}
task main()
{
int previousJoySlope = 0;
int TurnSpeed = 70;
while (true)
{
getJoystickSettings(joystick);
if (abs(joystick.joy1_x1) > 10 || abs(joystick.joy1_y1) > 10)
{
writeDebugStreamLine("Test successful.");
motor[DriveMotor] = calculateDrivePower(joystick.joy1_x1,joystick.joy1_y1);
}
}
}
If anyone could provide any insight, that'd be fantastic, thanks.

Answer 1

if (joyStickX != 0 & joyStickY != 0)
{
int joyDistance = (sqrt(pow((joyStickX - 0),2)+ pow((-joyStickY - 0),2)));
joyDistance = ((joyDistance/127)*100);
return (joyDistance);
}
The first issues that appears is the conditional within the if statement. The you have a single & that most likely should be &&:
if (joyStickX != 0 && joyStickY != 0)
(note: likely should be is used above because you can provide a conditional using a logical & of the tests joystickx != 0, but in this case it provides the same result. In that case, I would suggest the more readable && be used)
The next part of the code is simply the vector distance between 0,0 and the present position of the joystick. Of the general form dist^2 = (x2-x1)^2 + (y2-y1)^2 in your case x1,y1 = 0,0. Taking the square root of both sides provides dist = sqrt((x2-x1)^2 + (y2-y1)^2), or in C notation:
dist = (sqrt(pow((joyStickX - 0),2)+ pow((-joyStickY - 0),2)));
Next you have a scaling applied of dist = dist * 100/127 which provides the final distance returned. Hopefully this will help you understand what the code is doing.