Question

我正在寻找有关如何格式化二进制数的建议，以便每隔4位后就有一个空格。我有一个C程序，将十进制数转换为二进制数，但它只给出一个长字符串，没有像“10000000”这样的空格，我希望它是“1000 0000”

编辑：这是代码

1 #include "binary.h"
  2 
  3 char* binary(int num)
  4 {
  5    int i, d, count;
  6    char *pointer;
  7 
  8    count = 0;
  9    pointer = (char*)malloc(32+1);
 10 
 11    if(pointer == NULL)
 12      exit(EXIT_FAILURE);
 13 
 14    for (i = 31; i >= 0; i--)
 15    {
 16       d = num >> i;
 17 
 18        if (d & 1)
 19          *(pointer + count) = 1 + '0';
 20        else
 21          *(pointer + count) = 0 + '0';
 22 
 23        count++;
 24     }
 25     *(pointer+count) = '\0';
 26 
 27     return pointer;
 28 }

Answer 1

尝试以下更改：

将您的malloc更改为：

pointer = malloc(32+7+1); /* 32 digits + 7 spaces + null */

并在count++;：

之前将以下内容添加到循环中

/* if i is non-zero and a multiple of 4, add a space */
if (i && !(i & 3)) {
  count++;
  *(pointer + count) = ' ';
}

Answer 2

一种方法是将此二进制数转换为字符串变量，然后在每个第四和第五位之间放置一个空格。 Here是如何将int转换为字符串的提示。

Answer 3

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

void
printbin(unsigned v)
{
    size_t e = sizeof(v) * 10;
    char s[e+1];

    s[e--] = 0;

    for (; v || e % 5; v >>= 1) {
        if (e % 5 == 0) s[e--] = ' ';
        s[e--] = (v & 1) ? '1' : '0';
    }

    printf("%s\n", &s[e+1]);
}

Answer 4

#include <limits.h>

char* binary(int num){
    int i, count, bits = sizeof(num)*CHAR_BIT;
    char *pointer = malloc(bits + (bits / 4 -1) + 1);//for bit, for space, for EOS

    if(pointer == NULL){
        perror("malloc at binary");
        exit(EXIT_FAILURE);
    }

    count = 0;
    for (i=bits-1; i >= 0; i--){
        pointer[count++] = "01"[(num >> i) & 1];
        if(i && (count+1) % 5 == 0)
            pointer[count++] = ' ';
    }
    pointer[count] = '\0';

    return pointer;
}

格式化二进制数

4 个答案: