Extracting N-bit signed integer from a byte array

Well, in .NET for Windows, signed integers are stored in little endian two's complement format in any case so, as long as the last byte in the array is the most significant, then a four byte array will map directly to an Int32.

However, the three and five byte arrays will need some manipulation to find out what number they represent as there are no corresponding integral types in .NET.

Incidentally, the range you mentioned is for one's complement format. For two's complement the range is:

-2^(N-1) to +2^(N-1) – 1

Subject to all that I think the following should work though I'd test it thoroughly:

using System;

class Test

{

static void Main()

{

byte[] ba = {0, 0, 0, 128};

Console.WriteLine(GetInteger(ba));

byte[] ba2 = {255, 255, 255, 127};

Console.WriteLine(GetInteger(ba2));

Console.WriteLine();

byte[] ba3 = {0, 0, 128};

Console.WriteLine(GetInteger(ba3));

byte[] ba4 = {255, 255, 127};

Console.WriteLine(GetInteger(ba4));

Console.WriteLine();

byte[] ba5 = {0, 0, 0, 0, 128};

Console.WriteLine(GetInteger(ba5));

byte[] ba6 = {255, 255, 255, 255, 127};

Console.WriteLine(GetInteger(ba6));

Console.ReadKey();

}

static long GetInteger(byte[] ba)

{

switch(ba.Length)

{

case 4 :

return (long)BitConverter.ToInt32(ba, 0);

case 3 :

byte[] temp3 = {ba[0], ba[1], ba[2], 0};

if(temp3[2] >= 128) temp3[3] = 255;

return (long)BitConverter.ToInt32(temp3, 0);

case 5 :

byte[] temp5 = {ba[0], ba[1], ba[2], ba[3], ba[4], 0, 0, 0};

if(temp5[4] >= 128)

{

for(int i = 5; i < 8; i++) temp5[i] = 255;

}

return BitConverter.ToInt64(temp5, 0);

default :

throw new ArgumentException("Byte array must have length 3, 4 or 5");

}

}

The output which should represent the minimum and maximum values of the 4, 3 and 5 byte numbers is :

-2147483648

2147483647

-8388608

8388607

-549755813888

549755813887

1 Reply

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Vulpes13y ago
Well, in .NET for Windows, signed integers are stored in little endian two's complement format in any case so, as long as the last byte in the array is the most significant, then a four byte array will map directly to an Int32.

However, the three and five byte arrays will need some manipulation to find out what number they represent as there are no corresponding integral types in .NET.

Incidentally, the range you mentioned is for one's complement format. For two's complement the range is:

-2^(N-1) to +2^(N-1) – 1

Subject to all that I think the following should work though I'd test it thoroughly:

using System;

class Test
{
static void Main()
{
byte[] ba = {0, 0, 0, 128};
Console.WriteLine(GetInteger(ba));
byte[] ba2 = {255, 255, 255, 127};
Console.WriteLine(GetInteger(ba2));
Console.WriteLine();

byte[] ba3 = {0, 0, 128};
Console.WriteLine(GetInteger(ba3));
byte[] ba4 = {255, 255, 127};
Console.WriteLine(GetInteger(ba4));
Console.WriteLine();

byte[] ba5 = {0, 0, 0, 0, 128};
Console.WriteLine(GetInteger(ba5));
byte[] ba6 = {255, 255, 255, 255, 127};
Console.WriteLine(GetInteger(ba6));
Console.ReadKey();
}

static long GetInteger(byte[] ba)
{
switch(ba.Length)
{
case 4 :
return (long)BitConverter.ToInt32(ba, 0);

case 3 :
byte[] temp3 = {ba[0], ba[1], ba[2], 0};
if(temp3[2] >= 128) temp3[3] = 255;
return (long)BitConverter.ToInt32(temp3, 0);

case 5 :
byte[] temp5 = {ba[0], ba[1], ba[2], ba[3], ba[4], 0, 0, 0};
if(temp5[4] >= 128)
{
for(int i = 5; i < 8; i++) temp5[i] = 255;
}
return BitConverter.ToInt64(temp5, 0);

default :
throw new ArgumentException("Byte array must have length 3, 4 or 5");
}
}
}

The output which should represent the minimum and maximum values of the 4, 3 and 5 byte numbers is :

-2147483648
2147483647

-8388608
8388607

-549755813888
549755813887