Converting byte[] to array issue

Question

Please check this code

float f = BitConverter.ToSingle(new byte[] { 0xBF, 0x04, 0x8E, 0xFF }, 0);
byte[] b = BitConverter.GetBytes(f);

this yeilds a strange result. b will be { 0xBF, 0x04, 0xCE, 0xFF }

I guess it is because the value of f is NaN. the reason I'm asking this question is because I use Marsal to convert a stream of bytes into a struct that contains a float, the I swap endianity

The thing is when I get to the field is already messed up (like the above example)

any ideas?

Thanks!

I can't.I'm converting a byte[] to struct which it's inner structure is not known to me at that stage. — TheDude
– TheDude, Commented Nov 28, 2013 at 8:30
@harold if you use BitConverter in both directions, endianness isn't a concern - indeed, that would be the issue if the bytes were reversed, not if they are just mutated — Marc Gravell
– Marc Gravell, Commented Nov 28, 2013 at 8:31
@MarcGravell it is a concern this time, because the original input (ie { 0xBF, 0x04, 0x8E, 0xFF }) is the wrong way around. That makes a NaN, and then things start going south. — user555045
– user555045, Commented Nov 28, 2013 at 8:34
@TheDude please clarify, is this value supposed to be a NaN? Or something like -0.51780695 — user555045
– user555045, Commented Nov 28, 2013 at 9:20

AgentFire · Accepted Answer · 2013-12-03 07:25:26Z

2

Since your problem appears to an endian one, I'm wondering if checking for NaN and reversing the order and setting a boolean to reverse it back if necessary, would help:

byte[] input = { 0xBF, 0x04, 0x8E, 0xFF };
bool reversed = false;
float f = BitConverter.ToSingle(input, 0);

if (float.IsNaN(f))
{
    reversed = true;
    f = BitConverter.ToSingle(input.Reverse().ToArray(), 0);
}

byte[] b = BitConverter.GetBytes(f);

if (reversed)
    b = b.Reverse().ToArray();

edited Dec 3, 2013 at 7:25

AgentFire

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answered Nov 28, 2013 at 8:59

tinstaafl

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5 Comments

AgentFire Over a year ago

f == float.NaN will always be false. Change it to float.IsNaN(f).

Royi Namir Over a year ago

@AgentFire viva le Resharper

AgentFire Over a year ago

@RoyiNamir eh no, without any tools, I just checked the code out.

Royi Namir Over a year ago

@AgentFire , well , My resharper warned me about it.

Royi Namir Over a year ago

@AgentFire The best.(IMO).

AgentFire · Accepted Answer · 2013-11-28 08:57:32Z

0

This works:

double f = BitConverter.ToDouble(new byte[] { 0, 0, 0, 0, 0xBF, 0x04, 0x8E, 0xFF }, 0);
byte[] b = BitConverter.GetBytes(f);

This one also does:

int f = BitConverter.ToInt32(new byte[] { 0xBF, 0x04, 0x8E, 0xFF }, 0);
byte[] b = BitConverter.GetBytes(f);

I believe the reason is in the single floating-point storage. The third byte ({ 0xBF, 0x04, [0x8E to 0xBE] , 0xFF } is being +4'd for some reason while converted.

The good containter for 4 bytes is Int32 and I advise you to use it.

edited Nov 28, 2013 at 8:57

answered Nov 28, 2013 at 8:47

AgentFire

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3 Comments

Rik Over a year ago

That double represents a completely different number and is irrelevant. This is not a decimal precision issue, because the number is already in binary format. The problem here is that the bytes result in NaN. If OP just wanted to have a container for 4 bytes, why not just use a byte[]?

AgentFire Over a year ago

@Rik at least because byte[] as well as any other array is a reference typed object. And the double is still a container as well.

Craig.Feied Over a year ago

This should be the selected answer, because it not only explains clearly what caused the byte in question to be altered, but it also shows two good solutions when 4-byte data chunks must be passed through BitConverter twice in opposite directions, but they cannot be guaranteed to be valid single floats. There are a number of situations in which this arises. This was very helpful to me.

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