比較雙打工作不正常（總是顯示不同）

Question

我有一個程序來計算一個聚合區和周長，程序收到一個帶座標的文本文件並計算出面積。

我在計算中遇到了一些問題。現在我試圖比較雙打，我不明白爲什麼它不工作。

我有3行文本文件：

1.0 2.5 5.1 5.8 5.9 0.7 
1.2 4.1 5.1 5.8 6.8 1.9 2.9 0.2 
1.7 4.9 5.1 5.8 7.0 2.8 4.8 0.1 1.5 1.4 

我期待這個結果：

double expectedarea = 11.77;  
    double expectedperimeter = 15.64;  
    double expectedarea1 = 18.10;  
    double expectedperimeter1 = 17.02;  
    double expectedarea2 = 21.33;  
    double expectedperimeter2 = 16.60; 

所以我預計，消息「數組是相同的」出現了3因爲我給出正確的值，但我得到的消息數組是不同的3例。

你明白爲什麼我總是得到消息數組是不同的嗎？

#include <stdio.h> 
#include <stdlib.h> 
#include <math.h> 
#include <string.h> 
enum {x, y}; 
typedef struct triangle { 
    double v1[2]; 
    double v2[2]; 
    double v3[2]; 
} triangle; 
double area(triangle a); 
double perimeter(double *vertices, int size); 
double side(double *p1, double *p2); 

double expectedarea = 11.77;  
double expectedperimeter = 15.64;  
double expectedarea1 = 18.10;  
double expectedperimeter1 = 17.02;  
double expectedarea2 = 21.33;  
double expectedperimeter2 = 16.60;  
int main() 
{ 
    int idx; 
    int triangles; 
    int index; 
    int xycount; 
    double xy; 
    double triangle_area; 
    double polygon_area; 
    double perim; 
    double polygon_vertices[50] = {0.0}; 
    triangle a; 
    FILE* data; 
    char line[256]; 
    char* token; 
    if ((data = fopen("test.txt", "r")) == NULL) { 
     fprintf(stderr, "can't open data file\n"); 
     exit (EXIT_FAILURE); 
    } 
    while (fgets(line, sizeof (line), data)){ 
     xycount = 0; 
     polygon_area = 0; 
     line[strlen(line) - 1] = 0; 
     token = strtok(line, " "); 
     while (token != NULL){ 
      xy = atof(token); 
      token = strtok(NULL, " "); 
      polygon_vertices[xycount++] = xy; 
     } 
     idx = 0; 
     triangles = (xycount/2) - 2; 
     for (index = 2, idx = 0;idx < triangles;index += 2, ++idx){ 
      a.v1[x] = polygon_vertices[0]; 
      a.v1[y] = polygon_vertices[1]; 
      a.v2[x] = polygon_vertices[index + 0]; 
      a.v2[y] = polygon_vertices[index + 1]; 
      a.v3[x] = polygon_vertices[index + 2]; 
      a.v3[y] = polygon_vertices[index + 3]; 
      triangle_area = area(a); 
      polygon_area += triangle_area; 
     } 
     printf("area=%f\t", polygon_area); 
     perim = perimeter(polygon_vertices, xycount); 
     printf("perimeter=%f\n", perim); 

     if(polygon_area == expectedarea && perim == expectedperimeter) { 
      printf("Arrays are the same"); 
     } 
     if(polygon_area == expectedarea1 && perim == expectedperimeter1) { 
      printf("Arrays are the same"); 
     } 
     if(polygon_area == expectedarea2 && perim == expectedperimeter2) { 
      printf("Arrays are the same"); 
     } 

     else { 
      printf("Arrays are the different");  } 
    } 
    fclose(data); 
    return 0; 
} 
/* calculate triangle area with Heron's formula */ 
double area(triangle a) 
{ 
    double s1, s2, s3, S, area; 
    s1 = side(a.v1, a.v2); 
    s2 = side(a.v2, a.v3); 
    s3 = side(a.v3, a.v1); 
    S = (s1 + s2 + s3)/2; 
    area = sqrt(S*(S - s1)*(S - s2)*(S - s3)); 
    return area; 
} 
/* calculate polygon perimeter */ 
double perimeter(double *vertices, int size) 
{ 
    int idx, jdx; 
    double p1[2], p2[2], pfirst[2], plast[2]; 
    double perimeter; 
    perimeter = 0.0; 
    /* 1st vertex of the polygon */ 
    pfirst[x] = vertices[0]; 
    pfirst[y] = vertices[1]; 
    /* last vertex of polygon */ 
    plast[x] = vertices[size-2]; 
    plast[y] = vertices[size-1]; 
    /* calculate perimeter minus last side */ 
    for(idx = 0; idx <= size-3; idx += 2) 
    { 
     for(jdx = 0; jdx < 4; ++jdx) 
     { 
      p1[x] = vertices[idx]; 
      p1[y] = vertices[idx+1]; 
      p2[x] = vertices[idx+2]; 
      p2[y] = vertices[idx+3]; 
     } 
     perimeter += side(p1, p2); 
    } 
    /* add last side */ 
    perimeter += side(plast, pfirst); 
    return perimeter; 
} 
/* calculate length of side */ 
double side(double *p1, double *p2) 
{ 
    double s1, s2, s3; 
    s1 = (p1[x] - p2[x]); 
    s2 = (p1[y] - p2[y]); 
    s3 = (s1 * s1) + (s2 * s2); 
    return sqrt(s3); 
} 

Answer 1

在測試OP的代碼時，發佈的列表在0.7之後有一個空格。通常這個空格不是問題，但是我保存文件text.txt導致該行以" \r\n"結尾，而'\r'創建了額外的標記。額外的空白字符到strtok(NULL, " \n\r\t")解決了這個問題。

<1.0 2.5 5.1 5.8 5.9 0.7 > 
<1.2 4.1 5.1 5.8 6.8 1.9 2.9 0.2> 
<1.7 4.9 5.1 5.8 7.0 2.8 4.8 0.1 1.5 1.4> 

取決於列表是如何保存，最後一行可能會或可能不會與'\n'結束。由於OP可能會將最後一行顯示爲"1.7 4.9 5.1 5.8 7.0 2.8 4.8 0.1 1.5 1."（缺少最後一個4），所以這會成爲OP用下面的方法刪除'\n'的方法的問題。

line[strlen(line) - 1] = 0; 

更好地使用下面的，不需要最後'\n'正常工作。

line[strcspn(line, "\n")] = '\0'; 

---

OP的代碼是創建在其xy列表太多了座標。由於其他空白區域，它也有潛在的麻煩。添加更完整的列表。

while (token != NULL) { 
    xy = atof(token); 
    polygon_vertices[xycount++] = xy; 
    token = strtok(NULL, " \n\r\t"); // more white-spaces. 
    if (token == NULL) break; 
} 

代碼然後將計算出答案接近的預期值。在修復之後，應用先前提出的重複。 Is floating point math broken?

area=11.775000 expected area=11.770000 
    perimeter=15.645596 expected perimeter=15.640000 

而不是比較精確匹配FP號碼，代碼需要允許小的公差。請參閱comparing double values in C,
precision of comparing double values with EPSILON in C。

---

可以進行各種簡化和精度改進。例如：

#include <assert.h> 
/* calculate polygon perimeter */ 
double perimeter(double *vertices, int size) { 
    assert(size % 2 == 0 && size >= 0); // Insure only positive pairs are used 
    double perimeter = 0.0; 
    if (size > 1) { 
    double x_previous = vertices[size - 2]; 
    double y_previous = vertices[size - 2 + 1]; 
    while (size > 1) { 
     double x = *vertices++; 
     double y = *vertices++; 
     // hypot() certainly as accurate than sqrt(x*x + y*y) and avoids overflow 
     perimeter += hypot(x - x_previous, y - y_previous); 
     x_previous = x; 
     y_previous = y; 
     size -= 2; 
    } 
    } 
    return perimeter; 
}