浮点比较的奇怪结果 [英] Strange results with floating-point comparison

问题描述

我有这个简单的测试：

  double h; 
 ... 
 //分配h其初始值的代码，在下面使用
 ... 
 if（（h> 0）&&（h < ）{
 //分支1  - 计算
} 
 else {
 //分支2-无计算
} 
  

我列出了我的值，因为我有一些非常奇怪的结果，例如如果：
h = 1那么第一个分支到达，不明白为什么，因为如果h = 1我想要branch2计算。

我很困惑的东西这么明显？

---

编辑：

这是我计算然后使用 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]; 
 
 //计算差异
 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; 
  

}

---

编辑：

好吧，所以很清楚如何重新拟定条件。

From：

  double h; 
 if（h == 1）{
 //计算
} 
  

$ b b

To：

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

$ b

但是如何重新格式化这个？

  double h; 
 if（h <1）{
 //这里计算
} 
  

c>> c>是一个双精度值，它可能已经足够接近1，打印为 1 ，但实际上它小于1，因此比较成功。 浮点数很多。

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
}

解决方案

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.

这篇关于浮点比较的奇怪结果的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！