Convert array of unsigned char (byte) into unsigned short in C++

Question

I have array^ byteArray and I need to extract bytes in Little Endian sequence to make unsigned shorts and ints. I've tried every combination of the following I can think of so am asking for help.

int x = UInt32(byteArray[i]) + (UInt32)(0x00ff) * UInt32(byteArray[i + 1]);
int x = UInt32(byteArray[i]) + UInt32(0x00ff) * UInt32(byteArray[i + 1]);
int x = byteArray[i] + 0x00ff * byteArray[i + 1];

The problem is the least significant byte (at i+1) I know it is 0x50 but the generated short/int reports the lower byte as 0x0b. The higher byte is unaffected.

I figure this is a sign error but I can't seem to be able to fix it.

Hans Passant · Accepted Answer · 2011-09-24T23:03:19.463

3

You are using managed code. Endian-ness is an implementation detail that the framework is aware of:

array<Byte>^ arr = gcnew array<Byte> { 1, 2, 3, 4 };
int value = BitConverter::ToInt16(arr, 1);
System::Diagnostics::Debug::Assert(value == 0x302);

Whether the framework's assumptions are correct depends on where the data came from.

edited Sep 24 '11 at 23:03

answered Sep 24 '11 at 22:54

Hans Passant

922,412
146
1,693
2,536

On the MSDN help page I find ToInt16 sample indicating the array 01-02-03-04 yields ToInt16(arr, 1) == 0x20. Unless I missed something. Thanks for the speedy expertise offered. – John Sep 24 '11 at 23:30

K-ballo · Answer 2 · 2011-09-24T23:02:26.743

2

The proper way to generate an 16 bit int from two 8 bit ints is value = static_cast< int16_t >( hibyte ) << 8 | lobyte;

edited Sep 24 '11 at 23:02

answered Sep 24 '11 at 22:45

K-ballo

80,396
20
159
169

You should also cast `hibyte` to the 16-bit data type, otherwise, if it is an 8-bit data type, the left shift will turn it into a 0. – Praetorian Sep 24 '11 at 23:00

score 1 · Answer 3 · edited May 23 '17 at 12:03

1

int y = byteArray[i] | byteArray[i + 1] << 8;

is what you need to use. (see also Convert a vector<unsigned char> to vector<unsigned short>)

edited May 23 '17 at 12:03

Community

1
1

answered Sep 24 '11 at 22:45

John

6,433
7
47
82

score 0 · Answer 4 · answered Sep 24 '11 at 22:46

0

You want to do this instead

int x = UInt32(byteArray[i]) | (UInt32(byteArray[i + 1]) << 8);

Your multipliers are messing things up.

answered Sep 24 '11 at 22:46

Mark Robinson

3,135
1
22
37

You're trying to do a shift. The multiplication can emulate a shift but you're multiplying by 255 which won't emulate a shift. – Mark Robinson Sep 24 '11 at 23:13

score 0 · Answer 5 · answered Sep 24 '11 at 22:47

0

You have to multiply the second byte with 0x0100 instead of 0x00ff.

It's like in the decimal system, where you multiply by ten, not by nine.

answered Sep 24 '11 at 22:47

Roland Illig

40,703
10
88
121

Even so that doesn't account for making the byte 0x50 into 0x0b. – John Sep 24 '11 at 22:58

Convert array of unsigned char (byte) into unsigned short in C++

5 Answers5