Question

我正在评估，必须用C编写一个简短的程序来输入顶点和边的大小，以显示多边形的周长和面积。程序终止后，我必须显示：

a。计算总次数
b。所有顶点的总和
c。各边尺寸之和

这是如何用C完成的？谢谢

我尝试将它们存储在数组中，然后显示它们

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

int main() {
    float PI = 3.1415;
    char choice;
    float area, parameter;
    int radius, side, length, width;

    do {
        printf("Please enter a shape(C:Circle, S:Square, R:Rectangle, Q:Quiit> ");
        scanf("%s", &choice);

        switch (choice) {
          case 'C':
            printf("Enter a radius of the circle: ");
            scanf("%d", &radius);
            area = (2 * radius) * PI;
            parameter = 2 * PI * radius;
            printf("The area of the circle is %.02f and parameter is %.02f",
                   area, parameter);
            break;

          case 'S':
            printf("Enter the side of the square: ");
            scanf("%d", &side);
            area = side * side;
            parameter = 4 * side;
            printf("The area of the circle is %.02f and parameter is %.02f",
                   area, parameter);
            break;

          case 'R':
            printf("Enter the width of the rectangle: ");
            scanf("%d", &width);
            printf("Enter the length of the rectangle: ");
            scanf("%d", &length);
            area = length * width;
            parameter = (2 * length) + (2 * width);
            printf("The area of the circle is %.02f and parameter is %.02f",
                   area, parameter);
            break;

          case 'Q':
            printf ("Thank and bye");
            break;

          default:
            printf ("Invalid input");
        }
        return 0;
    } while (choice != 'Q');
}

我希望这是通过数组完成的，但是我不确定数组如何工作。

Answer 1

首先对当前代码做一些说明

拥有

 char choice;
 ...
 scanf ("%s", &choice);

您从 choice 中至少写了一个字符来放置空字符，行为是不确定的

您要输入字符吗

scanf (" %c", &choice);

%s之前的空格允许绕过分隔符（换行符/空格）

为什么要return 0;循环到“ Q”？删除它。

在所有打印品的末尾添加换行符，以将结果与问题分开（或者当然将“请输入形状...”替换为“ \ n请输入形状...”）它与上一个打印的同一行）

我鼓励您检查是否通过 scanf 输入了有效的内容，否则您不知道是否为scanf("%d", &radius);输入了整数，因此请检查 scanf 返回1

我希望这是通过数组完成的，但是我不确定数组如何工作。

您不需要数组，对于“ a”，“ b”，“ c”和“ d”，总和可以每次更新，对于“ e”每次更新乘积，最后在< em>正方形

无论如何，如果您想记住，则需要多个数组，每个主题一个，则有两种解决方案：

您使用静态大小的数组，在这种情况下，您必须限制输入数量，以免超出范围
您使用动态数组，先使用 malloc 然后 realloc 增大其大小

然后在结尾（'Q'）处根据数组内容计算所需的值

例如，根据预处理器标识符ARRAYS以两种方式管理“ a”和“ e”：

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

#ifdef ARRAYS
typedef struct Vector {
  float * values;
  size_t nValues;
} Vector;

// initialize the array, must be called because to add values
void init(Vector * v)
{
  v->values = malloc(0);
  v->nValues = 0;
}

// a a new value into the vector
void add(Vector * v, float f)
{
  v->values = realloc(v->values, (++v->nValues) * sizeof(float));
  v->values[v->nValues - 1] = f;
}

// print the content of the array
void pr(Vector * v, const char * what)
{
  printf("there are %d %s :", v->nValues, what);
  for (size_t i = 0; i != v->nValues; ++i)
    printf(" %.02f", v->values[i]);
  putchar('\n');
}
#endif

int main ()
{
  float PI = 3.1415;
  char choice;
  float area, parameter;
  int radius, side, length, width;
#ifdef ARRAYS
  Vector calculations, areas;

  init(&calculations);
  init(&areas);
#else
  int calcNumber = 0;
  float prodAreas = 1;
#endif

  do {
    printf("Please enter a shape(C:Circle, S:Square, R:Rectangle, Q:Quit> ");
    scanf (" %c", &choice);

    switch (choice) {
    case 'C':
      printf ("Enter a radius of the circle: ");
      if (scanf ("%d", &radius) != 1) {
        puts("invalid value");
        return -1;
      }
      area = (2 * radius) * PI;
      parameter = 2 * PI * radius;
      printf ("The area of the circle is %.02f and parameter is %.02f\n",
              area, parameter);
#ifdef ARRAYS
      add(&calculations, area);
      add(&areas, area);
      add(&calculations, parameter);
#else
      calcNumber += 2;
      prodAreas *= area;
#endif
      break;

    case 'S':
      printf ("Enter the side of the square: ");
      if (scanf ("%d", &side) != 1) {
        puts("invalid value");
        return -1;
      }
      area = side * side;
      parameter = 4 * side;
      printf ("The area of the circle is %.02f and parameter is %.02f\n",
              area, parameter);
#ifdef ARRAYS
      add(&calculations, area);
      add(&areas, area);
      add(&calculations, parameter);
#else
      calcNumber += 2;
      prodAreas *= area;
#endif
      break;

    case 'R':
      printf ("Enter the width of the rectangle: ");
     if ( scanf ("%d", &width) != 1) {
        puts("invalid value");
        return -1;
      }
      printf ("Enter the length of the rectangle: ");
      if (scanf ("%d", &length) != 1) {
        puts("invalid value");
        return -1;
      }
      area = length * width;
      parameter = (2 * length) + (2 * width);
      printf ("The area of the circle is %.02f and parameter is %.02f\n",
              area, parameter);
#ifdef ARRAYS
      add(&calculations, area);
      add(&areas, area);
      add(&calculations, parameter);
#else
      calcNumber += 2;
      prodAreas *= area;
#endif
      break;

    case 'Q':
      puts ("Thank and bye");
      break;

    default:
      puts ("Invalid input");
    }
  } while (choice != 'Q');

#ifdef ARRAYS
  pr(&calculations, "calculations");

  pr(&areas, "areas");

  float e = 1;

  for (size_t i = 0; i != areas.nValues; ++i)
    e *= areas.values[i];
  printf("square root of the product of all areas : %.02f\n", sqrt(e));
#else
  printf("there are %d calculations\n", calcNumber);
  printf("square root of the product of all areas : %.02f\n", sqrt(prodAreas));
#endif

  return 0;
}

使用数组进行编译和执行：

pi@raspberrypi:/tmp $ gcc -DARRAYS -pedantic -Wall  -Wextra c.c -lm
pi@raspberrypi:/tmp $ ./a.out
Please enter a shape(C:Circle, S:Square, R:Rectangle, Q:Quit> C
Enter a radius of the circle: 1
The area of the circle is 6.28 and parameter is 6.28
Please enter a shape(C:Circle, S:Square, R:Rectangle, Q:Quit> S
Enter the side of the square: 1
The area of the circle is 1.00 and parameter is 4.00
Please enter a shape(C:Circle, S:Square, R:Rectangle, Q:Quit> Q
Thank and bye
there are 4 calculations : 6.28 6.28 1.00 4.00
there are 2 areas : 6.28 1.00
square root of the product of all areas : 2.51
pi@raspberrypi:/tmp $

没有数组的编译和执行：

pi@raspberrypi:/tmp $ gcc -pedantic -Wall  -Wextra c.c -lm
pi@raspberrypi:/tmp $ ./a.out
Please enter a shape(C:Circle, S:Square, R:Rectangle, Q:Quit> C
Enter a radius of the circle: 1
The area of the circle is 6.28 and parameter is 6.28
Please enter a shape(C:Circle, S:Square, R:Rectangle, Q:Quit> S
Enter the side of the square: 1
The area of the circle is 1.00 and parameter is 4.00
Please enter a shape(C:Circle, S:Square, R:Rectangle, Q:Quit> Q
Thank and bye
there are 4 calculations
square root of the product of all areas : 2.51

我让你为'b''c'和'd'

如何在C中显示来自输入的总和和计算

1 个答案: