还有一种用二进制打印的方法:先转换整数。

要以二进制格式打印6，请将6改为110，然后打印“110”。

绕过char buf[]问题。 printf()格式说明符，标志和字段，如“%08lu”，“%*lX”仍然可用。 不仅是二进制(以2为基数)，这种方法还可以扩展到其他以16为基数的基数。 仅限于较小的整数值。

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

unsigned long char_to_bin10(char ch) {
  unsigned char uch = ch;
  unsigned long sum = 0;
  unsigned long power = 1;
  while (uch) {
    if (uch & 1) {
      sum += power;
      }
   power *= 10;
   uch /= 2;
  }
  return sum;
}

uint64_t uint16_to_bin16(uint16_t u) {
  uint64_t sum = 0;
  uint64_t power = 1;
  while (u) {
    if (u & 1) {
      sum += power;
      }
    power *= 16;
    u /= 2;
  }
  return sum;
}

void test(void) {
  printf("%lu\n", char_to_bin10(0xF1));
  // 11110001
  printf("%" PRIX64 "\n", uint16_to_bin16(0xF731));
  // 1111011100110001
}

#include <stdio.h>
#include <conio.h>

void main()
{
    clrscr();
    printf("Welcome\n\n\n");
    unsigned char x='A';
    char ch_array[8];
    for(int i=0; x!=0; i++)
    {
        ch_array[i] = x & 1;
        x = x >>1;
    }
    for(--i; i>=0; i--)
        printf("%d", ch_array[i]);

    getch();
}

void binario(int num) {
  for(int i=0;i<32;i++){
    (num&(1<i))? printf("1"):
        printf("0");
  }  
  printf("\n");
}

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

void DisplayBinary(unsigned int n)
{
    int l = sizeof(n) * 8;
    for (int i = l - 1 ; i >= 0; i--) {
        printf("%x", (n & (1 << i)) >> i);
    }
}

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');
    }
}