Welcome to ShenZhenJia Knowledge Sharing Community for programmer and developer-Open, Learning and Share
menu search
person
Welcome To Ask or Share your Answers For Others

Categories

I have this simple test:

double h;
...
// code that assigns h its initial value, used below
...
if ((h>0) && (h<1)){
 //branch 1 -some computations
}
else{
 //branch 2- no computations
}

I listed my values as I got some really strange results and for example if: h=1 then the first branch is reached and I do not understand why since if h=1 I want branch2 to be computed.
Am I getting confused by something so obvious?


Edit:

This is how I compute and then use h:

double* QSweep::findIntersection(edge_t edge1,edge_t edge2) {  
point_t p1=myPoints_[edge1[0]];
point_t p2=myPoints_[edge1[1]];
point_t p3=myPoints_[edge2[0]];
point_t p4=myPoints_[edge2[1]];

double xD1,yD1,xD2,yD2,xD3,yD3,xP,yP,h,denom;
double* pt=new double[3];
        
// calculate differences  
xD1=p2[0]-p1[0];  
xD2=p4[0]-p3[0];  
yD1=p2[1]-p1[1];  
yD2=p4[1]-p3[1];  
xD3=p1[0]-p3[0];  
yD3=p1[1]-p3[1];    

xP=-yD1;
yP=xD1;
denom=xD2*(-yD1)+yD2*xD1;
if (denom==0) {
    return NULL;
}
else{
h=(xD3*(-yD1)+yD3*xD1)/denom;
}
std::cout<<"h is"<<h<<endl;
if (h < 1) std::cout<<"no"<<endl;
else std::cout<<"yes"<<endl;
if (h==1) {
    return NULL;
}
else{
if ((h>0)&&(h<1)){
    pt[0]=p3[0]+xD2*h;  
    pt[1]=p3[1]+yD2*h;
    pt[2]=0.00;
}
else{
    return NULL;
}
}


return pt;  

}


Edit:

Okay, so it is clear how I should reformulate the condition.

From:

double h;
if (h==1){
   //computations here
}

To:

double h;
if (abs(h-1)<tolerance){
  //computations here
}

When I use double numbers.

But how do I reformulate this?

double h;
if (h<1){
   //computations here
}
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
thumb_up_alt 0 like thumb_down_alt 0 dislike
294 views
Welcome To Ask or Share your Answers For Others

1 Answer

Since h is a double, it may have been close enough to 1 to print as 1, but it is actually a bit less than 1 so the comparison succeeds. Floating-point numbers do that a lot.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
thumb_up_alt 0 like thumb_down_alt 0 dislike
Welcome to ShenZhenJia Knowledge Sharing Community for programmer and developer-Open, Learning and Share
...