c - 程序输出问题

Question

下面是我的程序，它根据给定的 (x,y) 坐标确定多边形的周长和面积，但我似乎得到了错误的输出，而且我不明白为什么。

输入是:

  3  12867  1.0 2.0  1.0 5.0  4.0 5.0
  5  15643  1.0 2.0  4.0 5.0  7.8 3.5  5.0 0.4  1.0 0.4

第一个条目是点数，第二个条目是多边形 ID，之后的任何内容都是一组坐标。

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

#define MAX_PTS 100
#define MAX_POLYS 100
#define END_INPUT 0

struct Point {
    double x, y;
};

double getDistance(struct Point a, struct Point b) {
    double distance;
    distance = sqrt((a.x - b.x) * (a.x - b.x) + (a.y-b.y) *(a.y-b.y));
    return distance;
}

double polygon_area(int length, double x[], double y[]) {
    double area = 0.0;
    for (int i = 0; i < length; ++i) {
       int j = (i + 1) % length;
       area += (x[i] * y[j] - x[j] * y[i]);
    }
area = area / 2;
area = (area > 0 ? area : -1 * area);

return (area);
}

int main(int argc, char *argv[]) {
    int npoints, poly_id;
    struct Point a, b;

    if(scanf("%d %d", &npoints, &poly_id)) {
        int iteration = 0;
        scanf("%lf %lf", &a.x, &a.y);
        struct Point initialPoint = a;
        double perimeter = 0;  // i start with 0 value of parameter.     
        for (iteration = 1; iteration < npoints; ++iteration) {
            scanf("%lf %lf", &b.x, &b.y);  // take input for new-point.
            perimeter += getDistance(a, b); // add the perimeter.
        // for next iteration, new-point would be first-point in getDistance
            a = b; 
        }
        // now complete the polygon with last-edge joining the last-point
        // with initial-point.
        perimeter += getDistance(a, initialPoint);

        printf("First polygon is %d\n", poly_id);
        printf("perimeter = %2.2lf m\n", perimeter);

        scanf("%d %d", &npoints, &poly_id);
        double x[MAX_PTS], y[MAX_PTS];
        double area = 0;
        for (iteration = 0; iteration < npoints; ++iteration) {
            scanf("%lf %lf", &(x[iteration]), &(y[iteration]));
        }
        area = polygon_area(npoints, x, y); 

        printf("First polygon is %d\n", poly_id);
        printf("area = %2.2lf m^2\n", area);

    } else if(scanf("%d", &npoints)==0) {
        exit(EXIT_SUCCESS);
    }

    return 0;
}

我不断得到的输出是:

First polygon is 12867
perimeter = 10.24 m
First polygon is 15643
area = 19.59 m^2

但我想要的输出是:

First polygon is 12867
perimeter = 10.24 m
First polygon is 12867
area = 4.50 m^2 

或者:

First polygon is 12867
perimeter = 10.24 m
area = 4.50 m^2

如果有人能指出我出错的地方，我将不胜感激。

Answer 1

如果你还没有解决这个问题，那么你的问题就很明显了。您读取第一个多边形的数据，然后计算周长:

    perimeter += getDistance(a, initialPoint);

    printf("First polygon is %d\n", poly_id);
    printf("perimeter = %2.2lf m\n", perimeter);

然后，令人费解的是，您在计算面积之前读取了第二个多边形的数据:

    scanf("%d %d", &npoints, &poly_id);
    double x[MAX_PTS], y[MAX_PTS];
    double area = 0;
<snip>
    area = polygon_area(npoints, x, y); 

您期望如何使用第二个多边形的数据来获取第一个多边形的面积有点令人困惑。

您需要做的是在读取下一个多边形的数据之前计算每个多边形的周长和面积。您通常会提示输入要处理的多边形数，而不是使用 if 代码块，然后执行以下操作:

for (i = 0; i < npolys; i++)
{
    <read data for poly>
    <calculate perimeter>
    <calculate area>
    <display results>
}

接下来，当您期望用户提供数据时，请提示输入。不要只是让用户看着闪烁的光标，想知道你的程序是否挂起等等。一个简单的 printf 询问点数和多边形 ID 就可以很好地工作。然后，输入每个 x/y 对的类似提示将大大有助于消除困惑。考虑到上述所有因素，您的代码可以重写为:

int main (void) {

    size_t npoints, poly_id;
    size_t npolys = 0;
    size_t it = 0;
    struct Point a, b;

    printf ("\nNumber of polygons to enter: ");
    scanf (" %zu", &npolys);

    for (it = 0; it < npolys; it++)
    {
        double x[MAX_PTS], y[MAX_PTS];
        double perimeter = 0;
        double area = 0;
        size_t iter = 0;

        printf ("\nEnter npoints & poly_id    : ");
        scanf("%zu %zu", &npoints, &poly_id);

        printf ("Enter the first point X & Y: ");
        scanf("%lf %lf", &a.x, &a.y);

        x[iter] = a.x;
        y[iter] = a.y;

        struct Point initialPoint = a;

        for (iter = 1; iter < npoints; ++iter) 
        {
            printf ("           next point X & Y: ");
            scanf("%lf %lf", &b.x, &b.y);   /* input for new-point. */

            x[iter] = b.x;
            y[iter] = b.y;

            perimeter += getDistance(a, b); /* add the perimeter.   */
            a = b;                          /* new-pt is first-pt   */
        }

        /* complete polygon joining the last-point with initial-point. */
        perimeter += getDistance (b, initialPoint);
        area = polygon_area (npoints, x, y); 

        printf("\nPolygon %zu is %zu\n", it, poly_id);
        printf("perimeter = %2.2lf m\n", perimeter);
        printf("     area = %2.2lf m^2\n", area);

    }

    return 0;
}

但是为什么不为您的代码编写一个输入例程来读取数据文件并消除容易出错的用户输入呢？这需要一点时间，但并不难。这样您就可以完全分离代码的输入和处理功能。

这使您能够真正专注于以逻辑方式布置代码的处理部分，而不是让处理逻辑充斥着用户输入。下面的示例说明了如何将输入与代码逻辑分开来帮助保持代码的整洁和可读性。在开始第一次计算之前，它将所有数据读取到结构数组中。

当您将代码分成逻辑函数时，维护任何单独的计算(例如周长或面积)只需调整单个函数的逻辑就可以了。请查看以下内容，如果您有疑问，请告诉我:

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

#define MAXPTS 100
#define MAXPOLYS 100

typedef struct point {
    double x, y;
} point;

typedef struct polygon {
    size_t sz;
    size_t id;
    point *vertex;
} polygon;

double get_distance (point a, point b);
double poly_perim (polygon a);
double polygon_area (polygon pg);
polygon *read_data (char *fn);

int main (int argc, char **argv)
{
    if (argc < 2 ) {
        fprintf (stderr, "error: insufficient input, usage: %s filename\n", argv[0]);
        return 1;
    }

    size_t it = 0;
    size_t idx = 0;
    polygon *pg = read_data (argv[1]);

    if (!pg) return 1;

    while (pg[idx].sz)
    {
        printf ("\n id: %zu   points: %zu  perimeter: %6.2lf  area: %6.2lf\n\n", 
                pg[idx].id, pg[idx].sz, poly_perim (pg[idx]), polygon_area (pg[idx]));

        for (it = 0; it < pg[idx].sz; it++)
            printf ("  %5.2lf  %5.2lf\n", pg[idx].vertex[it].x, pg[idx].vertex[it].y);

        idx++;
    }

    return 0;
}

double get_distance (point a, point b)
{
    double distance;
    distance = sqrt ((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
    return distance;
}

double poly_perim (polygon a)
{
    int i = 0;
    double perim = get_distance (a.vertex[0], a.vertex[a.sz -1]);

    for (i = 1; i < a.sz; i++)
        perim += get_distance (a.vertex[i-1], a.vertex[i]);

    return perim;
}

double polygon_area (polygon pg)
{
    double area = 0.0;
    int i = 0;

    for (i = 0; i < pg.sz; ++i) 
    {
        int j = (i + 1) % pg.sz;
        area += (pg.vertex[i].x * pg.vertex[j].y - pg.vertex[j].x * pg.vertex[i].y);
    }

    area /= 2.0;
    area = area > 0 ? area : -1 * area;

    return area;
}

polygon *read_data (char *fn)
{
    char *ln = NULL;
    size_t n = 0;
    size_t it = 0;
    size_t idx = 0;
    ssize_t nchr = 0;
    FILE *fp = NULL;
    polygon *pg = NULL;

    if (!(fp = fopen (fn, "r"))) {
        fprintf (stderr, "%s() error: file open failed '%s'.\n", __func__, fn);
        exit (EXIT_FAILURE);
    }

    if (!(pg = calloc (MAXPOLYS, sizeof *pg))) {
        fprintf (stderr, "%s() error: virtual memory allocation failed.\n", __func__);
        exit (EXIT_FAILURE);
    }

    while ((nchr = getline (&ln, &n, fp)) != -1)
    {
        char *p = ln;
        char *ep = NULL;
        long lnum = 0;
        double dnum = 0;

        errno = 0;
        lnum = strtol (p, &ep, 10);
        if (errno == 0 && (p != ep && lnum != 0))
            pg[idx].sz = (size_t)lnum;
        else {
            fprintf (stderr, "%s() error: file read failure '%s'.\n", __func__, fn);
            exit (EXIT_FAILURE);
        }

        p = ep;
        errno = 0;
        lnum = strtol (p, &ep, 10);
        if (errno == 0 && (p != ep && lnum != 0))
            pg[idx].id = (size_t)lnum;
        else {
            fprintf (stderr, "%s() error: file read failure '%s'.\n", __func__, fn);
            exit (EXIT_FAILURE);
        }

        pg[idx].vertex = calloc (pg[idx].sz, sizeof *(pg[idx].vertex));
        if (!pg[idx].vertex) {
            fprintf (stderr, "%s() error: virtual memory allocation failed.\n", __func__);
            exit (EXIT_FAILURE);
        }

        for (it = 0; it < pg[idx].sz; it++)
        {
            p = ep;
            errno = 0;
            dnum = strtod (p, &ep);
            if (errno == 0 && (p != ep && lnum != 0))
                pg[idx].vertex[it].x = dnum;
            else {
                fprintf (stderr, "%s() error: file read failure '%s'.\n", __func__, fn);
                exit (EXIT_FAILURE);
            }

            p = ep;
            errno = 0;
            dnum = strtod (p, &ep);
            if (errno == 0 && (p != ep && lnum != 0))
                pg[idx].vertex[it].y = dnum;
            else {
                fprintf (stderr, "%s() error: file read failure '%s'.\n", __func__, fn);
                exit (EXIT_FAILURE);
            }
        }
        idx++;
        if (idx == MAXPOLYS) {
            fprintf (stderr, "%s() warning: MAXPOLYS reached in file '%s'.\n", __func__, fn);
            break;
        }
    }

    fclose (fp);
    if (ln) free (ln);

    return pg;
}

输入

$ cat dat/poly.txt
  3  12867  1.0 2.0  1.0 5.0  4.0 5.0
  5  15643  1.0 2.0  4.0 5.0  7.8 3.5  5.0 0.4  1.0 0.4

输出

$ ./bin/poly dat/poly.txt

 id: 12867   points: 3  perimeter:  10.24  area:   4.50

   1.00   2.00
   1.00   5.00
   4.00   5.00

 id: 15643   points: 5  perimeter:  18.11  area:  19.59

   1.00   2.00
   4.00   5.00
   7.80   3.50
   5.00   0.40
   1.00   0.40