How do I turn a binary string into a float or doub

2019-02-26 11:04发布

In this question, Bill The Lizard asks how to display the binary representation of a float or double.

What I'd like to know is, given a binary string of the appropriate length, how could I perform the reverse operation (in C#)? In other words, how do I turn a binary string into a float or double?

As a side note, are there any bit strings which would not result in a valid float or double?


EDIT: By binary string I mean a string of 0s and 1s.

So, my input will be a string like this:

01010101010101010101010101010101

and my output should be a floating point number. (Or, if there were 64 bits in the string, a double.)

4条回答
混吃等死
3楼-- · 2019-02-26 11:53

Here's a solution that doesn't use BitConverter and isn't limited by the range of Int64.

static double BinaryStringToDouble(string s)
{
  if(string.IsNullOrEmpty(s))
    throw new ArgumentNullException("s");

  double sign = 1;
  int index = 1;
  if(s[0] == '-')
    sign = -1;
  else if(s[0] != '+')
    index = 0;

  double d = 0;
  for(int i = index; i < s.Length; i++)
  {
    char c = s[i];
    d *= 2;
    if(c == '1')
      d += 1;
    else if(c != '0')
      throw new FormatException();
  }

  return sign * d;
}

This version supports binary strings that represent values between to Double.MinValue and Double.MaxValue, or 1023 significant binary digits. It overflows to Double.PositiveInfinity or Double.NegativeInfinity.

@LukeH's answer only supports binary strings that represent values between to Int64.MinValue and Int64.MaxValue, or 63 significant binary digits.

Why you'd have a need for a binary string that is more than 63 digits in length is up for discussion.

If you don't want to allow a leading sign character, you can use this simpler version that only returns positive values.

static double BinaryStringToDouble(string s)
{
  if(string.IsNullOrEmpty(s))
    throw new ArgumentNullException("s");

  double d = 0;
  foreach(var c in s)
  {
    d *= 2;
    if(c == '1')
      d += 1;
    else if(c != '0')
      throw new FormatException();
  }

  return d;
}
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The star\"
4楼-- · 2019-02-26 11:56
string bstr = "01010101010101010101010101010101";
long v = 0;
for (int i = bstr.Length - 1; i >= 0; i--) v = (v << 1) + (bstr[i] - '0');
double d = BitConverter.ToDouble(BitConverter.GetBytes(v), 0);
// d = 1.41466386031414E-314
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放荡不羁爱自由
5楼-- · 2019-02-26 11:56
double d1 = 1234.5678;
string ds = DoubleToBinaryString(d1);
double d2 = BinaryStringToDouble(ds);

float f1 = 654.321f;
string fs = SingleToBinaryString(f1);
float f2 = BinaryStringToSingle(fs);

// ...

public static string DoubleToBinaryString(double d)
{
    return Convert.ToString(BitConverter.DoubleToInt64Bits(d), 2);
}

public static double BinaryStringToDouble(string s)
{
    return BitConverter.Int64BitsToDouble(Convert.ToInt64(s, 2));
}

public static string SingleToBinaryString(float f)
{
    byte[] b = BitConverter.GetBytes(f);
    int i = BitConverter.ToInt32(b, 0);
    return Convert.ToString(i, 2);
}

public static float BinaryStringToSingle(string s)
{
    int i = Convert.ToInt32(s, 2);
    byte[] b = BitConverter.GetBytes(i);
    return BitConverter.ToSingle(b, 0);
}
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