I think you are looking for something along the lines of a 7-bit encoded integer:
protected void Write7BitEncodedInt(int value)
{
uint num = (uint) value;
while (num >= 0x80)
{
this.Write((byte) (num | 0x80));
num = num >> 7;
}
this.Write((byte) num);
}
(taken from System.IO.BinaryWriter.Write(String)
).
The reverse is found in the System.IO.BinaryReader
class and looks something like this:
protected internal int Read7BitEncodedInt()
{
byte num3;
int num = 0;
int num2 = 0;
do
{
if (num2 == 0x23)
{
throw new FormatException(Environment.GetResourceString("Format_Bad7BitInt32"));
}
num3 = this.ReadByte();
num |= (num3 & 0x7f) << num2;
num2 += 7;
}
while ((num3 & 0x80) != 0);
return num;
}
I do hope this is not homework, even though is really smells like it.
EDIT:
Ok, so to put it all together for you:
using System;
using System.IO;
namespace EncodedNumbers
{
class Program
{
protected static void Write7BitEncodedInt(BinaryWriter bin, int value)
{
uint num = (uint)value;
while (num >= 0x80)
{
bin.Write((byte)(num | 0x80));
num = num >> 7;
}
bin.Write((byte)num);
}
static void Main(string[] args)
{
MemoryStream ms = new MemoryStream();
BinaryWriter bin = new BinaryWriter(ms);
for(int i = 1; i < 1000; i++)
{
Write7BitEncodedInt(bin, i);
}
byte[] data = ms.ToArray();
int size = data.Length;
Console.WriteLine("Total # of Bytes = " + size);
Console.ReadLine();
}
}
}
The total size I get is 1871 bytes for numbers 1-1000.
Btw, could you simply state whether or not this is homework? Obviously, we will still help either way. But we would much rather you try a little harder so you can actually learn for yourself.
EDIT #2:
If you want to just pack them in ignoring the ability to decode them back, you can do something like this:
protected static void WriteMinimumInt(BinaryWriter bin, int value)
{
byte[] bytes = BitConverter.GetBytes(value);
int skip = bytes.Length-1;
while (bytes[skip] == 0)
{
skip--;
}
for (int i = 0; i <= skip; i++)
{
bin.Write(bytes[i]);
}
}
This ignores any bytes that are zero (from MSB to LSB). So for 0-255 it will use one byte.
As states elsewhere, this will not allow you to decode the data back since the stream is now ambiguous. As a side note, this approach crams it down to 1743 bytes (as opposed to 1871 using 7-bit encoding).