如果是＆gt;一种是真的吗？ [英] When is a>a true?

问题描述

右键，我想我真的是生活在梦中。我有下面这段code的，我编译和AIX机器上运行：

  AIX 3 5
PowerPC_POWER5处理器类型
IBM XL C / C ++的AIX，V10.1
版本：10.01.0000.0003
＃包括LT＆;＆stdio.h中GT;
＃包括LT＆;＆math.h中GT;＃定义RADIAN（X）（（X）* ACOS（0.0）/ 90.0）双nearest_distance（双半径，双lon1，双LAT1，双lon2，双LAT2）{
    双rlat1 = RADIAN（LAT1）;
    双rlat2 = RADIAN（LAT2）;
    双rlon1 = lon1;
    双rlon2 = lon2;
    双A = 0，B = 0，C = 0;    A = SIN（rlat1）*罪（rlat2）+ COS（rlat1）* COS（rlat2）* COS（rlon2-rlon1）;
    的printf（％LF \\ n，一）;
    如果（一个大于1）{
      的printf（aaaaaaaaaaaaaaaa \\ n）;
    }
    B = ACOS（一）;
    C =半径* B;    返回半径*（ACOS（SIN（rlat1）*罪（rlat2）+
        COS（rlat1）* COS（rlat2）* COS（rlon2-rlon1）））;}INT主（INT ARGC，字符** argv的）{
  nearest_distance（6367.47,10,64,10,64）;
  返回0;
}
 

现在，值'一'的计算被报告为'1'后。而且，这AIX机器上，它看起来像1> 1作为我的如果输入的是真实的！而我的是什么，我认为是ACOS'1'返回NanQ因为1比1更大请问如何，甚至是可能的吗？我不知道该怎么想了！

在code适用于其他架构就好其中'一'真的需要什么，我认为是1，ACOS（a）为0的值。

解决方案

  

如果你做一个比较，其中的结果，expctedResult是浮动类型：

 如果（结果== expectedResult）
 

  

然后，它是不太可能的比较将是正确的。如果比较结果为真，那么它可能是不稳定的。 - 输入值，编译器，或CPU中的微小变化可能改变的结果，使该比较为假

有小量的比较 - 绝对误差

 如果（晶圆厂（结果 -  expectedResult）LT; 0.00001）
 

比较浮点数

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什么每个计算机科学家应该知道关于浮点运算

Right, I think I really am living a dream. I have the following piece of code which I compile and run on an AIX machine:

AIX 3 5
PowerPC_POWER5 processor type
IBM XL C/C++ for AIX, V10.1
Version: 10.01.0000.0003


#include <stdio.h>
#include <math.h>

#define RADIAN(x) ((x) * acos(0.0) / 90.0)

double nearest_distance(double radius,double lon1, double lat1, double lon2, double lat2){
    double rlat1=RADIAN(lat1);
    double rlat2=RADIAN(lat2);
    double rlon1=lon1;
    double rlon2=lon2;
    double a=0,b=0,c=0;

    a = sin(rlat1)*sin(rlat2)+ cos(rlat1)*cos(rlat2)*cos(rlon2-rlon1);
    printf("%lf\n",a);
    if (a > 1) {
      printf("aaaaaaaaaaaaaaaa\n");
    }
    b = acos(a);
    c = radius * b;

    return radius*(acos(sin(rlat1)*sin(rlat2)+
        cos(rlat1)*cos(rlat2)*cos(rlon2-rlon1)));

}

int main(int argc, char** argv) {
  nearest_distance(6367.47,10,64,10,64);
  return 0;
}

Now, the value of 'a' after the calculation is reported as being '1'. And, on this AIX machine, it looks like 1 > 1 is true as my 'if' is entered !!! And my acos of what I think is '1' returns NanQ since 1 is bigger than 1. May I ask how that is even possible ? I do not know what to think anymore !

The code works just fine on other architectures where 'a' really takes the value of what I think is 1 and acos(a) is 0.

解决方案

If you do a comparison where result and expctedResult are float types:

if (result == expectedResult)

Then it is unlikely that the comparison will be true. If the comparison is true then it is probably unstable – tiny changes in the input values, compiler, or CPU may change the result and make the comparison be false.

Comparing with epsilon – absolute error

if (fabs(result - expectedResult) < 0.00001)

From Comparing floating point numbers

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What Every Computer Scientist Should Know About Floating-Point Arithmetic

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