Convert double value in Binary

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hallo guy, please i got a problem when i try to Convert a double value in Binary

C#

public void DoubleinBinaereundHexa(double wert)
{

    long bitCount = sizeof(double) * 8;
    char[] result = new char[bitCount];


    long lgValue = BitConverter.ToInt64(BitConverter.GetBytes(wert), 0);


    for (long bit = 0; bit < bitCount; ++bit)
    {
        long maskwert = lgValue & (1 << bit);
        if (maskwert > 0)
        {
            maskwert = 1;
        }


        result[bitCount - bit -1] = maskwert.ToString()[0];
    }
    Console.Write("\n\nBinaere Darstellung:\t");

    for (int i = 0; i < 64; i++)
    {

        if (i % 4 == 0)
            Console.Write(" ");
        if (result[i] == '-')
        {
            result[i] = '1';
        }
        Console.Write(result[i]);

    }
}

Please i appologyse the fault ist in German language
Fehler 1 Der Operator '<<' kann nicht auf Operanden vom Typ 'int' und 'long' angewendet werden.

i think that it means in english: Operator '<<' cannot be applied to operands of type 'int' and 'long'

Please have somebody an Idea??

thx in advance

Posted 7-Nov-12 3:16am

stefan from germany

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Comments

__John_ 7-Nov-12 9:35am

I hope someone can answer this.
It seems the '<<' operator cannot be applyed to any type, short, int or long. Makes no sense.

__John_ 7-Nov-12 10:13am

Try changing 'bit' from 'long' to 'int'.

2 solutions

Solution 1

I think the problem is down to the numeric types you're using. Amend your left shift to:

C#

((long)1 << bit)

This is discussed in greater detail here:

http://social.msdn.microsoft.com/Forums/en/csharpgeneral/thread/1e9d6e3b-bbad-45df-9391-7403becd9641[^]

Update: My bad, looks like I've lost the ability to read somewhere today. Looking at the linked article, the right hand operand of a left shift is always an int:

Shift left:
int operator <<(int x, int count);
uint operator <<(uint x, int count);
long operator <<(long x, int count);
ulong operator <<(ulong x, int count);

Posted 7-Nov-12 3:30am

jim lahey

Updated 7-Nov-12 3:41am

v2

Comments

stefan from germany 7-Nov-12 9:34am

it doesn't walk i already try it Jim

jim lahey 7-Nov-12 9:42am

Please see update

Thomas Daniels 7-Nov-12 9:36am

If I try this, I'm getting error:
"Operator '<<' can't be applied to operands of type 'long' and 'long'".

stefan from germany 7-Nov-12 9:53am

Jim please can u explain me i don't understand correctly what it means. how can i use it?

jim lahey 7-Nov-12 9:58am

It means the number to the right of the "<<" needs to be an integer. the value of 'bit' either needs to be an integer or you need to rethink your bitwise shift

stefan from germany 7-Nov-12 11:42am

when i Change bit to int it run without problem but i get a wrong result

like this:

for (int bit = 0; bit < bitCount; ++bit)
{
long maskwert = lgValue & (1 << bit);
if (maskwert > 0)
{
maskwert = 1;
}

result[bitCount - bit -1] = maskwert.ToString()[0];
}
Console.Write("\n\nBinaere Darstellung:\t");

for (int i = 0; i < 64; i++)
{

if (i % 4 == 0)
Console.Write(" ");
if (result[i] == '-')
{
result[i] = '1';
}
Console.Write(result[i]);

}
}

/*
compiler run without fault but i get at the end a Wrong result

*/
For Example:
Double wert = 18, 4
/*
Wrong result ist :
1110 0000 0000 0000 0000 0000 0000 0000 1110 0000 0000 0000 0000 0000 0000 0000

Normaly a correct result muss looking like this:
0100 0000 0011 0010 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110

*/
please what i make wrong where is the Problem?

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Alan N · Accepted Answer · 2012-11-07T07:27:00

OK, lets take it apart and see what is wrong:

I'll modify the code to show the bytes of the double and all 64 bit masks.

C#

public void DoubleinBinaereundHexa(double wert) {
  int bitCount = sizeof(double) * 8;
  char[] result = new char[bitCount];

  //long lgValue = BitConverter.ToInt64(BitConverter.GetBytes(wert), 0);

  // split the conversion into two operations
  Byte[] bytes = BitConverter.GetBytes(wert);
  // show each byte
  foreach (Byte b in bytes) {
    Console.WriteLine(Convert.ToString(b, 2).PadLeft(8, '0'));
  }

  long lgValue = BitConverter.ToInt64(bytes, 0);

  for (int bit = 0; bit < bitCount; ++bit) {
    // show each mask
    Console.WriteLine(Convert.ToString((1 << bit), 2).PadLeft(64, '0'));

    long maskwert = lgValue & (1 << bit);
    if (maskwert > 0) {
      maskwert = 1;
    }
    result[bitCount - bit - 1] = maskwert.ToString()[0];
  }
  Console.WriteLine("\n\nBinaere Darstellung:");

  for (int i = 0; i < 64; i++) {

    if (i % 4 == 0)
      Console.Write(" ");
    if (result[i] == '-') {
      result[i] = '1';
    }
    Console.Write(result[i]);
  }
}

This shows that the bit masks produced by (1 << bit) are incorrect. In fact they are integer (32 bit) masks and (1 << 0) and (1 << 32) give the same mask. To correct this the literal value must be a long, i.e. (1L << bit).

With that modification the output is

0100 0000 0011 0010 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110

Alan.