Question

我想获得以下小数值

51043465443420856213到Arduino中的二进制文件。

结果应该是这个。

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

Answer 1

下面是一个代码模板，可以帮助您前进，因为我不想给您带来所有乐趣。

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

Answer 2

显然，您的值不会被存储为无符号长整数c程序，因为它大于64位。 Arduino有python。这是一个python函数，可以执行您想要的操作。该示例使用80位。如果你真的想要它们，你可以添加空格。

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：
00000000000000101100010001011110110100010110000110000100000001101101111110010101

Arduino中二进制的大十进制值

2 个答案: