java用浮点数和双打计算

Question

I have problem with java maths. I have following expression:

double tripp = (timeNow-previousTime)/(seconds*Sys.getTimerResolution())*trip;

Where I am trying to calculate rotating speed of my cube. Here is my full code of that function here I do that:

public float getTripLenghtToGo(int seconds, double trip) {
    long timeNow = Sys.getTime();
    System.out.println("RES: "+Sys.getTimerResolution());
    double tripp = (timeNow-previousTime)/(seconds*Sys.getTimerResolution())*trip;
    System.out.println("("+timeNow+"-"+previousTime+")/("+seconds+"*"+Sys.getTimerResolution()+")*"+trip+" = "+tripp);
    if(tripp != 0){
        previousTime = timeNow;
    }
    return (float)tripp;
}

I use OpenGL so it must return float. It returns float, but always zero.

Answer 1

timeNow is long type. The "upper" part of your division gives a long. The whole division will give a long. You need to cast it to double, or the decimal part will be lost.

For example, the output of:

long timeNow = 10;
long previousTime = 1;
double trip = 0.1d;
int seconds = 50;

double trippCast = ((double) (timeNow-previousTime))/((double)seconds)*trip;
System.out.println(trippCast);
double trippWithoutCast = (timeNow-previousTime)/seconds*trip;
System.out.println(trippWithoutCast);

is:

0.018
0.0

Answer 2

Two possibilities:

• Use System.nanoTime() - it gives you more precise time measurement
• use ((double)(timeNow-previousTime))/(seconds*Sys.getTimerResolution()) - if you divide a long with int, the result will be long. And that's 0L