/* Convert an int to it's binary representation */

char *int2bin(int num, int pad)
{
 char *str = malloc(sizeof(char) * (pad+1));
  if (str) {
   str[pad]='\0';
   while (--pad>=0) {
    str[pad] = num & 1 ? '1' : '0';
    num >>= 1;
   }
  } else {
   return "";
  }
 return str;
}

/* example usage */

printf("The number 5 in binary is %s", int2bin(5, 4));
/* "The number 5 in binary is 0101" */

也许有人会发现这个解决方案很有用:

void print_binary(int number, int num_digits) {
    int digit;
    for(digit = num_digits - 1; digit >= 0; digit--) {
        printf("%c", number & (1 << digit) ? '1' : '0');
    }
}

我的解决方案返回一个int，然后可以在printf中使用。它还可以以大端序或小端序返回位。

#include <stdio.h>
#include <stdint.h>

int binary(uint8_t i,int bigEndian)
{
    int j=0,m = bigEndian ? 1 : 10000000;
    while (i)
    {
        j+=m*(i%2);
        if (bigEndian) m*=10; else m/=10;
        i >>= 1;
    }
    return j;
}

int main()
{
    char buf[]="ABCDEF";
    printf("\nbig endian = ");
    for (int i=0; i<5; i++) printf("%08d ",binary(buf[i],1));
    printf("\nwee endian = ");
    for (int i=0; i<5; i++) printf("%08d ",binary(buf[i],0));
    getchar();
    return 0;
}

输出

big endian = 01000001 01000010 01000011 01000100 01000101 01000110
wee endian = 10000010 01000010 11000010 00100010 10100010 01100010

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。

至于我，我为此编写了一些通用代码

#include<stdio.h>
void int2bin(int n, int* bin, int* bin_size, const int  bits);

int main()
{
    char ch;
    ch = 'A';
    int binary[32];
    int binary_size = 0;
    
    int2bin(1324, binary, &binary_size, 32);
    for (int i = 0; i < 32; i++)
    {
        printf("%d  ", binary[i]);
    }
    
    
    return 0;
}

void int2bin(int n, int* bin,int *bin_size,const int  bits)
{
    int i = 0;
    int temp[64];
    for (int j = 0; j < 64; j++)
    {
        temp[j] = 0;
    }
    for (int l = 0; l < bits; l++)
    {
        bin[l] = 0;
    }

    while (n > 0)
    {
        temp[i] = n % 2;
        n = n / 2;
        i++;
    }
    *bin_size = i;

    //reverse modulus values
    for (int k = 0; k < *bin_size; k++)
    {
        bin[bits-*bin_size+k] = temp[*bin_size - 1 - k];
    }
}

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