A small utility function in C to do this while solving a bit manipulation problem. This goes over the string checking each set bit using a mask (1< void printStringAsBinary(char * input) { char * temp = input; int i = 7, j =0;; int inputLen = strlen(input); /* Go over the string, check first bit..bit by bit and print 1 or 0 **/ for (j = 0; j < inputLen; j++) { printf("\n"); while (i>=0) { if (*temp & (1 << i)) { printf("1"); } else { printf("0"); } i--; } temp = temp+1; i = 7; printf("\n"); } }

这里有一个快速的技巧来演示如何做你想做的事情。

#include <stdio.h>      /* printf */
#include <string.h>     /* strcat */
#include <stdlib.h>     /* strtol */

const char *byte_to_binary
(
    int x
)
{
    static char b[9];
    b[0] = '\0';

    int z;
    for (z = 128; z > 0; z >>= 1)
    {
        strcat(b, ((x & z) == z) ? "1" : "0");
    }

    return b;
}

int main
(
    void
)
{
    {
        /* binary string to int */

        char *tmp;
        char *b = "0101";

        printf("%d\n", strtol(b, &tmp, 2));
    }

    {
        /* byte to binary string */

        printf("%s\n", byte_to_binary(5));
    }
    
    return 0;
}

可能有点超时，但是如果你只是为了调试来理解或回溯你正在做的一些二进制操作而需要这个，你可以看看wcalc(一个简单的控制台计算器)。使用-b选项可以得到二进制输出。

e.g.

$ wcalc -b "(256 | 3) & 0xff"
 = 0b11

快速简单的解决方法:

void printbits(my_integer_type x)
{
    for(int i=sizeof(x)<<3; i; i--)
        putchar('0'+((x>>(i-1))&1));
}

适用于任何大小类型以及有符号整型和无符号整型。'&1'需要处理有符号整型，因为移位可能会进行符号扩展。

有很多方法可以做到这一点。这里有一个超级简单的方法，用于从有符号或无符号32位类型中打印32位或n位(如果有符号，则不输入负号，只打印实际的位)，并且不返回回车符。注意，i在移位前递减:

#define printbits_n(x,n) for (int i=n;i;i--,putchar('0'|(x>>i)&1))
#define printbits_32(x) printbits_n(x,32)

如果返回一个包含稍后存储或打印的比特的字符串呢?你可以分配内存并返回它，用户必须释放它，或者你返回一个静态字符串，但如果它再次被调用，或者被另一个线程调用，它会被破坏。两种方法显示:

char *int_to_bitstring_alloc(int x, int count)
{
    count = count<1 ? sizeof(x)*8 : count;
    char *pstr = malloc(count+1);
    for(int i = 0; i<count; i++)
        pstr[i] = '0' | ((x>>(count-1-i))&1);
    pstr[count]=0;
    return pstr;
}

#define BITSIZEOF(x)    (sizeof(x)*8)

char *int_to_bitstring_static(int x, int count)
{
    static char bitbuf[BITSIZEOF(x)+1];
    count = (count<1 || count>BITSIZEOF(x)) ? BITSIZEOF(x) : count;
    for(int i = 0; i<count; i++)
        bitbuf[i] = '0' | ((x>>(count-1-i))&1);
    bitbuf[count]=0;
    return bitbuf;
}

电话:

// memory allocated string returned which needs to be freed
char *pstr = int_to_bitstring_alloc(0x97e50ae6, 17);
printf("bits = 0b%s\n", pstr);
free(pstr);

// no free needed but you need to copy the string to save it somewhere else
char *pstr2 = int_to_bitstring_static(0x97e50ae6, 17);
printf("bits = 0b%s\n", pstr2);

C标准库中没有这样输出二进制的格式化函数。printf家族支持的所有格式操作都是针对人类可读的文本。

还有一种想法是将数字转换为十六进制格式，然后将每个十六进制密码解码为四个“位”(1和0)。Sprintf可以为我们做位操作:

const char* binary(int n) {
  static const char binnums[16][5] = { "0000","0001","0010","0011",
    "0100","0101","0110","0111","1000","1001","1010","1011","1100","1101","1110","1111" };
  static const char* hexnums = "0123456789abcdef";
  static char inbuffer[16], outbuffer[4*16];
  const char *i;
  sprintf(inbuffer,"%x",n); // hexadecimal n -> inbuffer
  for(i=inbuffer; *i!=0; ++i) { // for each hexadecimal cipher
    int d = strchr(hexnums,*i) - hexnums; // store its decimal value to d
    char* o = outbuffer+(i-inbuffer)*4; // shift four characters in outbuffer
    sprintf(o,"%s",binnums[d]); // place binary value of d there
  }
  return strchr(outbuffer,'1'); // omit leading zeros
}

puts(binary(42)); // outputs 101010