7

我正在尝试使用以下算法将十进制数转换为 C 中的二进制数。我不明白为什么它对于某些输入不能正常工作（例如，对于 1993 年，我得到 1420076519）。

int aux=x;
long bin=0;
while (aux>0)
{
    bin=bin*10+aux%2;
    aux=aux/2;
}
printf("%d in decimal is %ld in binary.", x, bin);
4

7 回答 7

4

当您打印 long 时，您不会打印二进制文件。转换为二进制或显示十进制数的二进制表示的最佳方法是将其存储在字符串中。波纹管是另一个 SO答案中提供的解决方案

void getBin(int num, char *str)
{
  *(str+5) = '\0';
  int mask = 0x10 << 1;
  while(mask >>= 1)
    *str++ = !!(mask & num) + '0';
}
于 2012-11-02T18:33:03.207 回答
3

如果您知道算法，就没有理由不使用itoa

http://www.cplusplus.com/reference/clibrary/cstdlib/itoa/

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

int main ()
{
  int n;
  char output[100];

  printf("Enter a number: ");
  scanf("%d", &n);

  itoa(n, output, 2); //2 means base two, you can put any other number here

  printf("The number %d is %s in binary.", n, output);

  return 0;
}
于 2012-11-02T23:36:42.713 回答
2

转换是如何工作的？

/* Example: 
   125(10) -----> ?(2)                     125  |_2
                                            -1-   62  |_2
                                                  -0-   31 |_2
                                                        -1-  15 |_2
                                                             -1-  7 |_2
                                                                 -1-  3 |_2
                                                                     -1-  1 */

所以在这个例子中，125(10) 的二进制数是 1111101(2)，这就是我在函数中描述的过程。

/* Functions declaration (Prototype) */

 int wordCalculator( int * const word, long int number, int base );
    int main( void )
        {
            int i, base;
            int word[ 32 ];
            unsigned long int number;

            printf( "Enter the decimal number to be converted: " );
            scanf( "%ld", &number );
            printf( "\nEnter the new base: " );
            scanf( "%d", &base );

            i = wordCalculator( word, number, base );

            printf( "The number is: " );

            for(; i >= 0; i--){

                if ( word[ i ] <= 9)
                    printf( "%d", word[ i ] );

                else
                    /* 65 represents A in ASCII code. */
                    printf( "%c", ( 65 - 10 + word[ i ] ) );
            }

            printf( "\n" );
        }

        int wordCalculator( int * const word, long int number, int base )
        {
            unsigned long int result = number;
            int i, difference;

            i = 0;
            do{
                difference = result % base;
                result /= base;
                *( word + i ) = difference;
                i++;

                if ( result < base )
                    *( word + i ) = result;

            } while( result >= base );

            return i;

        }
于 2012-11-02T19:05:39.063 回答
0

我认为最短的答案是

char* getBinary(int n,char *s)
{
  while(n>0)
  {
    *s=(n&1)+'0';
    s++;
    n>>=1;
  }
  *s='\0';
  return s;
}

在被调用的函数中以相反的方式打印它..因为存储已经完成LSB但是MSB 我们必须先打印MSB然后LSB

于 2012-11-02T18:44:37.147 回答
0

您应该使用字符串来存储二进制数。以下代码应该适合您。

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

char *decimal_to_binary(int);

main()
{
   int n, c, k;
   char *pointer;

   printf("Enter an integer in decimal number system\n");
   scanf("%d",&n);

   pointer = decimal_to_binary(n);
   printf("Binary string of %d is: %s\n", n, pointer);

   free(pointer);

   return 0;
}

char *decimal_to_binary(int n)
{
   int c, d, count;
   char *pointer;

   count = 0;
   pointer = (char*)malloc(32+1);

   if ( pointer == NULL )
      exit(EXIT_FAILURE);

   for ( c = 31 ; c >= 0 ; c-- )
   {
      d = n >> c;

      if ( d & 1 )
         *(pointer+count) = 1 + '0';
      else
         *(pointer+count) = 0 + '0';

      count++;
   }
   *(pointer+count) = '\0';

   return  pointer;
}
于 2012-11-02T18:34:42.207 回答
0

您可以使用以下算法将十进制数转换为二进制数系统。

#include <stdio.h>  

int main()  
{  
    long long decimal, tempDecimal, binary;  
    int rem, place = 1;  

    binary = 0;  

    /* 
     * Reads decimal number from user 
     */  
    printf("Enter any decimal number: ");  
    scanf("%lld", &decimal);  
    tempDecimal = decimal;  

    /* 
     * Converts the decimal number to binary number 
     */  
    while(tempDecimal!=0)  
    {  
        rem = tempDecimal % 2;  

        binary = (rem * place) + binary;  

        tempDecimal /= 2;  
        place *= 10;  
    }  

    printf("\nDecimal number = %lld\n", decimal);  
    printf("Binary number = %lld", binary);  

    return 0;  
}
于 2015-08-24T06:33:49.143 回答
0

这是我写的一个递归解决方案，它很简单并且工作正常。

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

int  printBinary(int N)
{
    if(N < 0){errno = EINVAL; return -1;}

    if(N == 0)
        printf("0");
    else if(N == 1)
        printf("1");
    else
    {
        printBinary(N/2);
        printf("%d", N%2);
    }

    return 0;
}

int main(int argc, char* argv[])
{
    if(argc < 2)
    {
        fprintf(stderr, "usage: %s NUM\nWhere NUM is an integer number\n", argv[0]);
        exit(EXIT_FAILURE);
    }
    errno = 0;


    long NUM = strtol(argv[1], NULL, 10);
    if(NUM == 0 && errno != 0)
    {
        perror("Error during number acquisition: ");
        exit(EXIT_FAILURE);
    }

    if(NUM < 0)
    {
        printf("-");
        printBinary(-NUM);
    }
    else
        printBinary(NUM);

    printf("\n");

    return 0;
}
于 2019-03-20T19:37:55.460 回答