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[Reply #30 on:]

this is what i have, what should i change?

Code:

int AM9111::read(unsigned int address)
{
  int rv = 0;
  _read() = rv;
  _setAddress(address);
  return rv;
}

Code:

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

[Reply #31 on:]

Code:

_read() = rv;

I thought we'd already decided that this was a no-no?

Code:

int AM9111::read(unsigned int address)
{
  int rv = 0;
  _read() = rv;
  _setAddress(address);
  return rv;
}

Is that really the correct order of events?

[Reply #32 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;
}

Someone's been doing too much Visual Basic.

Instead of:

Code:

 _read() = _rbits;
  return rv;

You want:

Code:

 return _rbits;

Although I'm not sure what rv does. Did you put it there to make compiler errors go away?

[Reply #33 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;
}

I still don't totally get this. Are you trying to change _rbits? In which case why return anything from the function? In that case you want:

Code:

void AM9111::_read()
{
  digitalWrite(_odp,0);
  digitalWrite(_wep,1);
  _rbits = 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]));
}

But if you want to return stuff, and not affect things such as _rbits (have side-effects) then you actually want:

Code:

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

[Reply #34 on:]

Thanks, ill try it

[Reply #35 on:]

Yes! It finally compiled! I will upload it tomorrow morning and try it.

Thanks for all your help

And what do you mean by "Someone's been doing too much Visual Basic"?

[Reply #36 on:]

Well, have you done Visual Basic?

In VB, a function returns its results by assigning to the function name. For example:

Code:

function foo (a, b)
  foo = a + b   ' return result of adding a to b
end function

So when I saw you write:

Code:

_read() = _rbits;

I thought you might have thought this was how you returned the function value.

[Reply #37 on:]

Well i have done a little visual basic, but not much, i did more visual c++.

and i have made an arduino sketch thad did, for example:

Code:

boolean yn(int pin)
{
  yn = digitalRead(pin);
}

[Reply #38 on:]

Not one that compiled, right?

Code:

sketch_apr23c.cpp: In function 'boolean yn(int)':
sketch_apr23c:2: error: assignment of function 'boolean yn(int)'
sketch_apr23c:2: error: cannot convert 'int' to 'boolean ()(int)' in assignment

[Reply #39 on:]

hmm, well i know it was something like that, mabey i used if statements or something, i don't know. it was a while ago. Well ill test the lib tomorrow...

[Reply #40 on:]

Now the _read() always returns 0

here is the code:

Code:

#include "AM9111.h"
#include "WProgram.h"

/////////////////////////////////////////
//
// PUBLIC
//
AM9111::AM9111(uint8_t outputDisablePin, uint8_t writeEnablePin)
{
  _odp = outputDisablePin;
  _wep = writeEnablePin;
}

int AM9111::setIO(uint8_t IO0, uint8_t IO1, uint8_t IO2, uint8_t IO3)
{
  _IO[0] = IO0;
  _IO[1] = IO1;
  _IO[2] = IO2;
  _IO[3] = IO3;
}
 

int AM9111::read(unsigned int address)
{
  int rv = 0;
  _setAddress(address);
  return rv;
}

int AM9111::write(unsigned int address, uint8_t value)
{
  int rv = 0;
  _setAddress(address);
  rv = _write(value);
  return rv;
}

/////////////////////////////////////////
//
// PRIVATE
//
void AM9111::_setAddress(unsigned int address)
{
for(int p=0; p<8; p++)
{
  pinMode(p, OUTPUT);
  digitalWrite(p, bitRead(address, p));
}
}


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


int AM9111::_write(uint8_t value)
{
  digitalWrite(_odp,1);
  digitalWrite(_wep,0);
  int  rv = 0;
  digitalWrite(_IO[0],bitRead(0,value));
  digitalWrite(_IO[1],bitRead(1,value));
  digitalWrite(_IO[2],bitRead(2,value));
  digitalWrite(_IO[3],bitRead(3,value));
  return rv;
}

[Reply #41 on:]

is it because of the first read statement rv = 0?

[Reply #42 on:]

Yes.
That, and the fact that you don't call _read from read.

[Reply #43 on:]

how do you call _read from read?

[Reply #44 on:]

In your own reply #30, you very nearly did.
You just got it the wrong way round.