Question

所以我的一个朋友在使用具有功能的结构时，让他的代码正确地吐出一个边界（当你在＆＃34; x y＆＃34;形式中输入一个点）时会遇到一些问题。

编辑：代码应该占4分，使用欧几里德函数计算每一边。（你会在print_Perimeter函数中看到尝试）

此外，该程序应该在您点击“＆n”之后退出，但他无法弄明白，是否有人可以给出任何建议？

我试图给他一些帮助，但也无法自己解决。很抱歉，如果代码看起来有点被屠杀，我们尝试了一些事情并保存原样。

如果有任何需要更好的说明，我会尽力回复。提前谢谢！

typedef struct Point
  {
int xCoord;
int yCoord;

}point;

int point_Print(int x, int y)
  {
printf("(X Y): %d %d\n", x, y);
}

double print_TwoPoints(int a, int b, int c, int d)
  {
double answer = (sqrt(pow((a - c),2)) + (pow((b - d),2)));
printf("The distance between point (%d,%d) and point (%d,%d) is %.2lf.\n", a, b, c, d, answer);
}

void print_Perimeter(Point *p1, Point *p3, Point *p4)
  {
int perimeter = 2 * sqrt(pow((p1->xCoord - p4->xCoord),2) + pow((p1->yCoord - p4->yCoord), 2))
        + sqrt(pow((p1->xCoord - p3->xCoord),2) + pow((p1->yCoord - p3->yCoord), 2))
printf("The perimeter of the rectangle is: %d.\n", perimeter);
}

int main()
  {
char quitKey = 'y';
//supposed to make the program quit
while(quitKey != 'n')

  {

point xCoord1;
point yCoord1;
printf("Please enter values for first point, X1 Y1: \n");
scanf("%d %d", &(xCoord1.xCoord), &(yCoord1.yCoord));

point xCoord2;
point yCoord2;
printf("Please enter values for second point, X1 Y1: \n");
scanf("%d %d", &(xCoord2.xCoord), &(yCoord2.yCoord));

point xCoord3;
point yCoord3;
printf("Please enter values for third point, X1 Y1: \n");
scanf("%d %d", &(xCoord3.xCoord), &(yCoord3.yCoord));

point xCoord4;
point yCoord4;
printf("Please enter values for forth point, X1 Y1: \n");
scanf("%d %d", &(xCoord4.xCoord), &(yCoord4.yCoord));

point_Print(xCoord1.xCoord, yCoord1.yCoord);
point_Print(xCoord2.xCoord, yCoord2.yCoord);
point_Print(xCoord3.xCoord, yCoord3.yCoord);
point_Print(xCoord4.xCoord, yCoord4.yCoord);

print_TwoPoints(xCoord1.xCoord, yCoord1.yCoord, xCoord2.xCoord, yCoord2.yCoord);
print_TwoPoints(xCoord3.xCoord, yCoord3.yCoord, xCoord4.xCoord, yCoord4.yCoord);

Point p1;
p1.xCoord = xCoord1;
p1.yCoord = yCoord1;

Point p3;
p3.xCoord = xCoord3;
p3.yCoord = yCoord3;

Point p4;
p4.xCoord = xCoord4;
p4.yCoord = yCoord4;

print_Perimeter(&p1, &p3, &p4);
printf("Enter 'y' to enter new coordinates or enter 'n' to quit the program: \n");
scanf(" %c", &quitKey);

}
}

Answer 1

以下代码：

干净地编译
很直接
仅适用于矩形，而不适用于平行四边形
，并且point2比point1更高且更右。所以你需要测试/调整其他条件和其他3个象限
检查错误

现在是代码

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

typedef struct Point
{
    int x;
    int y;
} point;


void print_Perimeter( point *point1, point *point2 )
{
    int horizontal = point2->x - point1->x;
    int vertical   = point2->y - point1->y; // EDIT: corrected from '+' to '-'

    int perimeter  = 2*(horizontal+vertical);
    printf("The perimeter of the rectangle is: %d.\n", perimeter);
}


int main( void )
{
    int quitKey = 'y';

    //supposed to make the program quit
    while( quitKey != 'n')
    {
        point point1;
        printf("Please enter values for first point, X1 Y1: \n");
        if( 2 != scanf("%d %d", &point1.x, &point1.y) )
        {
            fprintf( stderr, "scanf for point1 coordinates failed\n");
            exit( EXIT_FAILURE );
        }

        // implied else, scanf successful

        point point2;
        printf("Please enter values for second point, X2 Y2: \n");
        if( 2 != scanf("%d %d", &point2.x, &point2.y) )
        {
            fprintf( stderr, "scanf for point2 coordinates failed\n");
            exit( EXIT_FAILURE );
        }

        // implied else, scanf successful

        print_Perimeter( &point1, &point2 );

        // clear stdin
        int ch;
        while( (ch = getchar()) != EOF && '\n' != ch );

        printf("%s\n", "Enter 'y' to enter new coordinates or enter 'n' to quit the program:");
        quitKey = getchar();
        quitKey = tolower( quitKey );
    }

}

使用C中的euclidean计算矩形的周长

1 个答案: