Search bits in an Array of bytes

Hi, I am receiving signals in bytes. The length of the signal is not the same, so in every signal the amount of bytes is not the same. So I have an array of:

byte x = Data[0]; // Data[0] different in every signal
for (byte y = 1, y<=x; y++)

Data[1]
Data[2]
.
.
Data[y] with “y” different in every signal and between 1 to 6.

Then i must look for specific combination of bits in every byte, and i don’t know how to do it.
Please, if someone can help me ?
Thank you!

Not sure what you mean, but would bitwise AND do the trick?

Every bit pattern is also going to have a byte value, so you could test for that, too.

Thank you for your answers,

Could you give me an example?
I am not in a good level yet.

Do you think it will work ?

byte x = Data[0]; // Data[0] different in every signal
for (byte y = 1, y<=x; y++)
{
Data = Data[y]
}

If (bitRead(data,7)==0
{

}

That won't work because the bitRead() will always be checking Data unless that is what you want.

If you can give us a better idea of what values are being searched for, we can probably give you a better code example.

Let me give you an example:

Signal 1 : 11001001 00110110 11100000 00110011 // ( Data[0] = 4)
// (Data[1] = 11001001, Data[2] = 00110110, Data[3] = 11100000, Data[4] = 00110011 )

then following Signal 2: 00000011 00011101 // ( Data[0] = 2)
// (Data[1] = 00000011, Data[2] = 00011101)

From every byte, decoding the first three bits, bit7,6,5 show me what to do.

This is what i want to do.

I don't know how to separate these bits.

Bitwise AND with 0xe0 (aka 0b11100000)

OK. Can you explain it more, please ?

How about the plain old:

if(Variable[0]==0xE0)

?

Otherwise I like to do:

if((Variable[0] && 0xE0) == 0xE0)

Bit if the bits in between can change, and that doesn’t matter:

if((Variable[0] & 0xE0) == 0xE0)

Ok. I think i understand it, but how can i check it in every byte ?

if((Variable[0] && 0xE0) == 0xE0)

You like it all you want, but it is WRONG.
Think about it; the result of (Variable [ 0 ] && 0xE0) will be 'true' (aka one) if Variable [ 0 ] is non-zero, and zero if Variable [ 0 ] is zero.
One or zero is never, ever going to be equal to 0xE0.

Sorry, but i can't understand everything. I am trying to learn.

You mean to substitute Variable[0], with Data[n] and add 0xE0 (or the combination i need to check).
If it is true, i am ok.

Ok, now i understand it.

Probably one of the simplest methods for you to use, that will be easy for you to follow would be

byte special=data[0]/32;

Since byte type variables can't hold fractions, all the least significant bits have simply been disposed of.
You will now have a byte value that holds anything from 0 to 7 (dependent on those three bits you were interested in)

If you want the three bits separately, e.g. each as booleans

boolean argA = Data[0] & 0x80 == 0x80; // 8th bit
boolean argB = Data[0] & 0x40 == 0x40; // 7th bit
boolean argC = Data[0] & 0x20 == 0x20; // 6th bit

Each argument will be true or false depending on whether those specific bits were set in Data[0], so you can handle them independently.

The specific bits can be checked by using bitwise AND against the value of the single bit.

0x80 = 128
0x40 = 64
0x20 = 32
0x10 = 16
0x08 = 8
0x04 = 4
0x02 = 2
0x01 = 1

Use those values to check whether or not specific bits are set.

Thank you for your answers,

One more question,

Data[0] is the number of the bytes in the signal.
If Data[1] starts from 11 then i must decode the bytes from Data[3] till [n-1]
If Data[1] starts from 00 then i must decode the bytes from Data[2] till [n-1]

I am thinking to use:

If ((Data[1] && 0xC0) == 0xC0)
{for (n=3, n<Data[0],n++)

if((Data[n] && 0xE0) == 0xE0)

}else{
for (n=2, n<Data[0],n++)
if((Data[n] && 0xE0) == 0xE0)

Is it correct?

If ((Data[1] && 0xC0) == 0xC0)

No! See my earlier comment.
Single &

You want & (bitwise AND) not && (logical AND)

Also, for loops use the following syntax

for (initialization; condition; dec/increment)

Note the use of semicolons as opposed to commas.

Yes, Yes,

You are correct, i wrote it wrong.
Single &, and ; instead of ,.