C# getting 1st 2 integers out of a byte array

Question

I have a byte[] like this

byte[] buffer = new byte[1024];

this byte[] may have values like this:

buffer = {0, 0, 0, 106, 0, 0, 0, 11, 64, 33, 50, 32, 32, 32, ....}

I am trying to get first 8 bytes, that is:

0,0,0,106

0,0,0,11

, and convert these into integers which is 106 and 11.

I can safely assume that first 8 bytes always represent 2 integers like in example above, these are 106 and 11 and that they take form of 4 bytes with 1st 3 being 0's like above.

Both are 4 byte signed integers in Hi-Lo order

How do I do that in C#?

@GiladGreen - only when the triple zeroes are also guaranteed. — H H, Feb 20 '19 at 17:59
Are we assuming these are 4-byte integers, most significant bytes first? — Joe Sewell, Feb 20 '19 at 18:02
.net has the `BinaryReader` https://learn.microsoft.com/en-us/dotnet/api/system.io.binaryreader?view=netframework-4.7.2 — Daniel A. White, Feb 20 '19 at 18:04
@DanielA.White - I would expect the same endiannes as BitConverter. — H H, Feb 20 '19 at 18:06
@DanielA.White, BinaryReader reads integers in little-endian format. OP specifically mentions that the integer octets/bytes are in Hi-Lo order (big-endian order) — , Feb 20 '19 at 18:17
@elgonzo i linked this in my answer. https://stackoverflow.com/questions/8620885/c-sharp-binary-reader-in-big-endian — Daniel A. White, Feb 20 '19 at 18:18
Yeah, just saw your answer now. My apologies for my superfluous comment :-D — , Feb 20 '19 at 18:19

Callum Watkins · Answer 1 · 2019-02-20T18:23:45.557

3

All you need is to access indices 3 and 7:

int first = buffer[3];
int second = buffer[7];

There is an implicit conversion from byte to int.

This is possible due to the following:

I can safely assume that [...] they take form of 4 bytes with 1st 3 being 0's

Therefore you only need the last byte of each 4-byte integer.

edited Feb 20 '19 at 18:23

answered Feb 20 '19 at 18:14

Callum Watkins

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@user6f6e65 yes, what did I miss? – Callum Watkins Feb 20 '19 at 18:15
He wants the first 4 bytes to be combined to make a 32-bit int. *Not* casting the bytes – user6f6e65 Feb 20 '19 at 18:16
1

@user6f6e65 Since the first three bytes will always be zero, only the last byte is needed from each integer. – Callum Watkins Feb 20 '19 at 18:17
2

it was ironic as i hadn't read the question thoroughly. lmao. you get my upvote – user6f6e65 Feb 20 '19 at 18:18
1

@user6f6e65 no worries haha – Callum Watkins Feb 20 '19 at 18:19
"with 1st 3 being 0's" was added later. – H H Feb 21 '19 at 06:30

score 3 · Accepted Answer · answered Feb 20 '19 at 18:14

A simple DIY function:

int BytesToInt32(byte[] buff, int offset)
{
    return (buff[offset + 0] << 24)
         + (buff[offset + 1] << 16)
         + (buff[offset + 2] << 8)
         + (buff[offset + 3]);
}

and then:

buffer = {0, 0, 0, 106, 0, 0, 0, 11, 64, 33, 50, 32, 32, 32, ....};
int a = BytesToInt32(buffer, 0);
int b = BytesToInt32(buffer, 4);

score 2 · Answer 3 · answered Feb 20 '19 at 18:07

2

I would convert your byte[] into a MemoryStream (or leave it as a Stream). Then use a BinaryReader as appropriate. If the endianness isn't correct: C# - Binary reader in Big Endian?

answered Feb 20 '19 at 18:07

Daniel A. White

187,200
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score 1 · Answer 4 · answered Feb 20 '19 at 18:09

1

private static int BytesToInt(byte[] array, int startIndex){
    int toReturn = 0;
    for (int i = startIndex; i < startIndex + 4; i++)
    {
        toReturn = toReturn << 8;
        toReturn = toReturn + array[i];
    }
    return toReturn;
}

answered Feb 20 '19 at 18:09

Joe Sewell

6,067
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score 0 · Answer 5 · answered Feb 20 '19 at 18:06

Use the convert class.

int myint1 = Convert.ToInt32(buffer[someIndex1]);
int myint2 = Convert.ToInt32(buffer[someIndex2]);

As others mentioned, if it's guaranteed that index 4 and index 7 has the non-zero bytes then you can just plug it in straight.

C# getting 1st 2 integers out of a byte array

5 Answers5