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[Reply #15 on:]

_read() = _rbits;

[Reply #16 on:]

What about it?

[Reply #17 on:]

_read isn't a function returning a reference, so the assignment isn't possible.

[Reply #18 on:]

OK, so how would I make it so you could do a statement like: Serial.print(AM9111.read()); ?

[Reply #19 on:]

Code:

int AM9111::_read()
{
  digitalWrite(_odp,0);
  digitalWrite(_wep,1);
  int rv = 0;
  bitWrite(_rbits,0,digitalRead(_IO[0]));
  bitWrite(_rbits,1,digitalRead(_IO[1]));
  bitWrite(_rbits,2,digitalRead(_IO[2]));
  bitWrite(_rbits,3,digitalRead(_IO[3]));
  _read() = _rbits;
  return rv;
}

Have you ever thought why this routine would always return zero? (if it were to compile)
Is that what you want?

[Reply #20 on:]

WHat i wanted was to have _read read the outputs of the ram and have _read = the output ex: out = 1101; _read = 13;

[Reply #21 on:]

If that's what you wanted, why does it return zero?

Your _write method also (unusually) returns a value that is always zero.

[Reply #22 on:]

I changed, but still has an error, do i have to have it do: return _rbits; ?

[Reply #23 on:]

Did you initialise _rbits?

[Reply #24 on:]

how?

as in int _rbits?

[Reply #25 on:]

No, apart from the missing semicolon, that looks more like a declaration, not an initialisation.

The clue is, you did initialise a variable, but it wasn't _rbits.

[Reply #26 on:]

so, how should i do it?

[Reply #27 on:]

I'd probably forget about _rbits altogether, and use (I mean really use) "rv".

[Reply #28 on:]

ok, ill try it.

[Reply #29 on:]

By the way, the spelling is "indeterminate", not "intermediate".    ;-)