C++ reference byte array - c++

Question

I have an issue with C++ and creating a reference byte[].
In C# my method is:
public static void SetBitAt(ref byte[] Buffer, int Pos, int Bit, bool Value)
{
byte[] Mask = { 0x01, 0x02, 0x04, 0x08, 0x10, 0x20, 0x40, 0x80 };
if (Bit < 0) Bit = 0;
if (Bit > 7) Bit = 7;
if (Value)
Buffer[Pos] = (byte)(Buffer[Pos] | Mask[Bit]);
else
Buffer[Pos] = (byte)(Buffer[Pos] & ~Mask[Bit]);
}
I want to translate it to C++, but I can't get the refworking for C++. I saw something about the & symbol and I tried something like this:
void SetBitAt(byte& buffer[], int Pos, int Bit, bool Value)
{
byte Mask[] = { 0x01, 0x02, 0x04, 0x08, 0x10, 0x20, 0x40, 0x80 };
if (Bit < 0) Bit = 0;
if (Bit > 7) Bit = 7;
if (Value)
{
buffer[Pos] = (byte)(buffer[Pos] | Mask[Bit]);
}
else
{
buffer[Pos] = (byte)(buffer[Pos] & ~Mask[Bit]);
}
}
but then I get the Error:
'buffer': arrays of references are illegal.
So how can I change my C++ code to work with a reference array?
EDIT:
I use this method for setting a buffer, but it doesn't change when I use this method.
other class:
buffer = ReadDB(2); //Read the values in the DataBlock
SetBitAt(buffer, 0,0 true); //Set bit 0,0 to 1(true)
WriteDB(2, buffer); //Write the values to the Datablock
but the buffer doesn't change. its the same value.

Answer 1

If you want to pass array by reference, you should
void SetBitAt(byte (buffer&)[10], int Pos, int Bit, bool Value)
But in your case, you don't need that, just
void SetBitAt(byte buffer[], int Pos, int Bit, bool Value)
Note in this case array will decay to pointer (i.e. byte*), that means the size of array won't be reserved as pass by reference would.

Answer 2

'buffer': arrays of references are illegal.
This is due to operator precedence. Saying byte &buffer[] is an array of references, while saying byte (&buffer)[size] is a reference to an array.
See C++ pass an array by reference for more details.
So how can I change my C++ code to work with a reference array?
When passing your array as a function argument, you should drop & symbol. You can still modify the contents of your array because the array's address is passed instead.
Assuming you have a typedef of char to byte, your function signature should look like this:
void SetBitAt(byte buffer[], int Pos, int Bit, bool Value) { ... }
Note that the above is equivalent to passing a pointer:
void SetBitAt(byte *buffer, int Pos, int Bit, bool Value) { ... }
Modifying the contents of your array is still a matter of saying buffer[Pos] = // some value;
This post on What is array decaying? should be useful.

Answer 3

Shouldn't it simply be like this:
void SetBitAt(byte buffer[], int Pos, int Bit, bool Value)
{
byte Mask[] = { 0x01, 0x02, 0x04, 0x08, 0x10, 0x20, 0x40, 0x80 };
if (Bit < 0) Bit = 0;
if (Bit > 7) Bit = 7;
if (Value)
{
buffer[Pos] = (byte)(buffer[Pos] | Mask[Bit]);
}
else
{
buffer[Pos] = (byte)(buffer[Pos] & ~Mask[Bit]);
}
}
In this way, buffer is passed as a pointer, and buffer[Pos] references to the Pos-th element of buffer. It is plain C, but it should work.

Answer 4

You can pass it simply by address as:
void SetBitAt(byte* buffer, int Pos, int Bit, bool Value) { ... }
or simply as:
void SetBitAt(byte buffer[], int Pos, int Bit, bool Value) { ... }
Either one will tell the compiler that byte pointer is passed to function, althrough with second header you omit pointer arithmetics ;)