Moving array variables

I'm reading a string of characters from serial but I'm only interested in one byte of data. I thought the most efficient way to do this was to use an array of 3 chars (char c[2] and look for my byte based on the bytes either side...

while (Serial.available() > 0) {
  c[0] = Serial.read();
  if (c[2] == 'P' && c[0] == 'V') {
    if (bitRead(c[1],0) == 0) {
      // do some stuff here
      }
    }
  c[2] = c[1];
  c[1] = c[0];
}

My code doesn't work. The last two lines are breaking it but I can't see what I'm doing wrong. I'm sure it's something basic.

Please help.

Thank you.

you're missing a } at the very end

Sorry, no, that was just be copy/pasting the bit of code out the bigger program. I've added it back into the top post, thanks.

There is a whole load of debug lines in my code now as I've been trying to work out what's happening :~

I thought the most efficient way to do this was to use an array of 3 chars (char c[2] and look for my byte based on the bytes either side...

An array of 2 characters, one of which will be the NULL, you mean. The value in the brackets is the count, not the upper index. You need to allow room for the trailing NULL, too.

You absolute star! Thank you

Why should there have to be a NULL, Paul? I see no string or print commands.

Dogsbody:
... use an array of 3 chars (char c[2] and look for my byte based on the bytes either side...

  if (c[2] == 'P' && c[0] == 'V') {

}

Regardless of whether there is a null or not, an array of 2 characters has indexes 0 and 1. So you can't check index 2 for a P.

Dogsbody:
.. to use an array of 3 chars (char c[2]

In fact, even that should give you pause. An array of 3 chars you say?

char c [3];

Isn't an initialization step missing ? Something like "if numCharsReceived >=3 else ".

I've changed this to the following now...

char c[3] = {0,0,0};

... is that what you mean?

If your array "c" is a global or a static, the compiler will helpfully set all elements to zero, unless you tell it to set them to some other value.
Difficult to tell from snippets though (hint)

Well, I guess explicitly initializing the array is a simpler way to solve the issue.