寻找素因数分解的程序

Question

我编写这段代码是为了找到一个数的素数分解。我只是无法弄清楚最后一部分。如果 x 作为 double 或 float 输入，则程序应打印错误消息并终止。我如何做到这一点？

#include <stdio.h>
int main()
{
int x, i;
printf("Enter an integer:  ");
scanf("%d", &x);

 if (x <= 1)
{
    return 1;
}
printf("The prime factorization of %d is ", x);
if (x > 1)
{
    while (x%2 == 0) 
    { 
        printf("2 "); 
        x = x / 2; 
    } 
     for ( i = 3; i < 1009; i = i + 2)
    {
    while (x%i == 0) 
        { 
            printf("%d ",i); 
            x = x / i; 
        }
    }
}
return 0;
}

Answer 1

您的起点应该涵盖所有需要和不需要的情况，因此您应该从用户那里获取float 号码，而不是int。然后，您应该检查数字的整个小数部分是否为0。也就是说，如果它们都等于0，则用户希望提供int数字而不是float。

第一步是声明一个浮点数：

float y;

之后，从用户那里获取它的值：

scanf("%f", &y);

第二步是检查是int还是float。这一步有很多方法。例如，我发现 roundf() function 很有用。它接受一个float 数字并计算最接近该数字的整数。因此，如果最接近的整数是数字本身，则该数字必须是int。对吧？

if(roundf(y)!=y)

如果你确定是int，你可以进行第三步：将float类型转换为int类型。它被称为type-casting。对于程序的其余部分，此步骤是必需的，因为在您的算法中，您使用 int 类型操作数字，因此只需将其转换为 int：

x = (int)y;

添加上面的行后，您可以使用您键入的其余代码。我给整个程序：

#include <stdio.h>
#include <math.h>
int main()
{
    int x,i;
    float y;
    printf("Enter an integer:  ");
    scanf("%f", &y);

    if(roundf(y)!=y){
        printf("%f is not an integer!",y);
        return 1;
    }

    else{
        x = (int)y;
    }

    if (x <= 1)
    {
        printf("%d <= 1",x);
        return 1;
    }

    else
    {
        printf("The prime factorization of %d is ", x);
        while (x%2 == 0)
        {
            printf("2 ");
            x = x / 2;
        }
        for ( i = 3; i < 1009; i = i + 2)
        {
            while (x%i == 0)
            {
                printf("%d ",i);
                x = x / i;
            }
        }
    }
    return 0;
}

Answer 2

scanf() 的使用有点棘手，我会不惜一切代价避免使用它来扫描用户生成的输入。但是这里是关于如何获取scanf() 错误的简短概述

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

int main(void)
{
   int x, i, scanf_return;

   printf("Enter an integer:  ");

   /* Reset "errno". Not necessary here, just in case. */
   errno = 0;
   /* scanf() returns a value in case of an error */
   scanf_return = scanf("%d", &x);
   /*
    * scanf() returns "EOF" if it didn't find all what you wanted or
    * and error happened.
    * It sets "errno" to the value of the actual error. See manpage
    * for all of the details.
    */
   if (scanf_return == EOF) {
      /*
       * The error is connected to the stream, so we can differ between
       * an error within scanf() and and error with the input stream
       * (here: stdin)
       */
      if (ferror(stdin)) {
         fprintf(stderr, "Something went wrong while reading stdin: %s\n", strerror(errno));
         exit(EXIT_FAILURE);
      } else {
         /* e.g. a conversion error, a float instead of an integer, letters
            instead of a decimal number */
         fprintf(stderr, "Something went wrong within scanf()\n");
         exit(EXIT_FAILURE);
      }
   }
   /*
    * If no error occurred, the return holds the number of objects
    * scanf() was able to read. We only need one, but it would throw an
    * error if cannot find any objects, so the check is here for
    * pedagogical reasons only.
    */
   if (scanf_return != 1) {
      fprintf(stderr, "Something went wrong within scanf(): wrong number of objects read.\n");
      exit(EXIT_FAILURE);
   }

   if (x <= 1) {
      fprintf(stderr, "Input must be larger than 1!\n");
      exit(EXIT_FAILURE);
   }
   printf("The prime factorization of %d is ", x);
   /* No need for that test, x is already larger than one at this point. */
   /* if (x > 1) { */
   while (x%2 == 0) {
      printf("2 ");
      x = x / 2;
   }
   for (i = 3; i < 1009; i = i + 2) {
      while (x%i == 0) {
         printf("%d ",i);
         x = x / i;
      }
   }
   /* } */
   /* Make it pretty. */
   putchar('\n');
   exit(EXIT_SUCCESS);
}

有效吗？

$ ./stackoverflow_003 
Enter an integer:  1234
The prime factorization of 1234 is 2 617 
$ factor 1234
1234: 2 617
$ ./stackoverflow_003 
Enter an integer:  asd
Something went wrong within scanf(): wrong number of objects read.
$ ./stackoverflow_003 
Enter an integer:  123.123
The prime factorization of 123 is 3 41

不，它不起作用。为什么不？如果您要求scanf() 扫描一个整数，它会抓取所有连续的十进制数字（0-9），直到没有一个。小限定符“连续”很可能是您的问题的根源：带有小数部分的浮点数有一个小数点，这就是 scanf() 假定您想要的整数结束的点。检查：

 $ ./stackoverflow_003 
 Enter an integer:  .123
 Something went wrong within scanf(): wrong number of objects read

你是怎么知道的？ @weather-vane 提供了多种方法之一：检查整数之后的下一个字符是否是句点（或您选择的另一个小数分隔符）：

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

int main(void)
{
   int x, i, scanf_return;
   char c = -1;

   printf("Enter an integer:  ");

   /* Reset "errno". Not necessary here, just in case. */
   errno = 0;
   /* scanf() returns a value in case of an error */
   scanf_return = scanf("%d%c", &x, &c);
   /*
    * scanf() returns "EOF" if it didn't find all what you wanted or
    * and error happened.
    * It sets "errno" to the value of the actual error. See manpage
    * for all of the details.
    */
   if (scanf_return == EOF) {
      /*
       * The error is connected to the stream, so we can differ between
       * an error within scanf() and and error with the input stream
       * (here: stdin)
       */
      if (ferror(stdin)) {
         fprintf(stderr, "Something went wrong while reading stdin: %s\n", strerror(errno));
         exit(EXIT_FAILURE);
      } else {
         /* e.g. a conversion error, a float instead of an integer, letters
            instead of a decimal number */
         fprintf(stderr, "Something went wrong within scanf()\n");
         exit(EXIT_FAILURE);
      }
   }
   /*
    * If no error occurred, the return holds the number of objects
    * scanf() was able to read. We can use this information now.
    * If there is a period (actually any character) after the integer
    * it returns 2 (assuming no error happened, of course)
    */

   /* If no integer given, the following character ("%c") gets ignored. */
   if (scanf_return == 0) {
      fprintf(stderr, "Something went wrong within scanf(): no objects read.\n");
      exit(EXIT_FAILURE);
   }
   /* Found two objects, check second one which is the character. */
   if (scanf_return == 2) {
      if (c == '.') {
         fprintf(stderr, "Floating point numbers are not allowed.\n");
         exit(EXIT_FAILURE);
      }
   }

   if (x <= 1) {
      fprintf(stderr, "Input must be larger than 1!\n");
      exit(EXIT_FAILURE);
   }
   printf("The prime factorization of %d is ", x);
   /* No need for that test, x is already larger than one at this point. */
   /* if (x > 1) { */
   while (x%2 == 0) {
      printf("2 ");
      x = x / 2;
   }
   for (i = 3; i < 1009; i = i + 2) {
      while (x%i == 0) {
         printf("%d ",i);
         x = x / i;
      }
   }
   /* } */
   /* Make it pretty. */
   putchar('\n');
   exit(EXIT_SUCCESS);
}

检查：

$ ./stackoverflow_003 
Enter an integer:  123
The prime factorization of 123 is 3 41 
$ ./stackoverflow_003 
Enter an integer:  123.123
Floating point numbers are not allowed.
$ ./stackoverflow_003 
Enter an integer:  .123
Something went wrong within scanf(): no objects read.

对我来说已经足够好了。有一个小错误：

$ ./stackoverflow_003 
Enter an integer:  123.
Floating point numbers are not allowed

但我想我可以把它留给亲爱的读者作为练习。