在Arduino中从大十进制值转换为二进制

我正在尝试获取以下十进制值

51043465443420856213到Arduino中的二进制

结果应该是这样

10 1100 0100 0101 1110 1101 0001 0110 0001 1000 0100 0000 0110 1101 1111 1001 0101

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显然，您的值不是以无符号长的c程序形式存储的，因为它大于64位。Arduino有python。下面是一个python函数，它可以实现您想要的功能。该示例使用80位。如果确实需要，可以添加空格

def int2bin(n, count=24):
    """returns the binary of integer n, using count number of digits"""
    return "".join([str((n >> y) & 1) for y in range(count-1, -1, -1)])

print int2bin(51043465443420856213, 80)  

输出：

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00000000000000 10110001011110110110100001011000011000000001101101111110010101下面是一个代码模板，我不想把所有的乐趣都浪费掉

void print_bin(char *s) {
  // Keep track of carry from previous digit - what should this be initialized to?
  int carry = ...;

  // Divide the "number string" by 2, one digit/character at a time 
  // for each array element in `s` starting with the first character 
  for (... ; ... ; ...) {
    // Extract the string element (character), covert to a decimal digit & add 10x carry
    int digit = ... 
    // Divide digit by 2 and then save as a character
    array element = ...
    // Mod the digit by 2 as the new carry
    carry = ...
  }
  // If first digit is a `0`, then move `s` to the next position.
  if (*s == '0') {
    ...
  }

  // If `s` is not at the end, recursively call this routine on s
  if (...) {
    print_bin(s);
  }

  // Since the more significant digits have been printed, now print
  // this digit (which is "mod 2" of the original "number string") as a character
  fputc('0' + carry, stdout);
}

int main(void) {
  char s[] = "51043465443420856213";
  print_bin(s);
}

结果

101100010001011110110100010110000110000100000001101101111110010101

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是否希望它以字符串形式出现，以便可以在1和0上进行迭代？二进制只是一个基数，所以从技术上讲，这个数字已经等同于二进制中的数字。结果是一个66位的二进制数。正在寻求什么：文本10 1100 0100。。。1001 0101或某些值为10 1100 0100的数字。。。10010101基地2？