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[on:]

I'm learning how to read and write data using the wire interface at the moment, but as my question says I'm unsure if the code is interpretting the address as a hex value or decimal, for example:

Code:

void setup()
{
  Wire.begin();        // join i2c bus (address optional for master)
  Serial.begin(9600);  // start serial for output
 
  writeTo(DEVICE, 0x16, 8);     
}  // end of setup

Where writeTO is:

Code:

void writeTo(int device, byte address, byte val)
  {
  Wire.beginTransmission(device); //start transmission to device
  Wire.write(address);        // send register address
  Wire.write(val);        // send value to write
  Wire.endTransmission(); //end transmission
  } // end of writeTo

So, i'm using writeTo(DEVICE, 0x16, , how does it know that the 0x16 is hex and not decimal?

[Reply #1 on:]

The '0x' part tells the compiler that you are using a hex constant.

[Reply #2 on:]

0x16 tells the code it is hex (the 0x part).
0x00 is a byte (8 bits) as in
byte flag1 = 0x01;

0x0000 is a int or word (16 bits), as in
int analogValue = analogRead (A0);

0x01234567 is a long (32 bits), as in
unsigned long currentTime = millis();

B11110000, B indicates binary.

'Nothing' preceding a number indicates it is decimal.

[Reply #3 on:]

B11110000, B indicates binary.

This is something the Arduino IDE does. There is also 0b11110000.

A leading 0 indicates an octal constant, I believe.

[Reply #4 on:]

'Nothing' preceding a number indicates it is decimal.

...and a leading zero indicates it is octal

[Reply #5 on:]

Thank you, it seems obvious now. If I had the register map below:



and I wanted to write FS_SEL=3  on register 0x16 then am I correct in saying:
FS_SEL =3 makes the line equal in binary:

00011000

which is equal to 24 in decimal, so i'd want:
writeTo(DEVICE, 0x16, 24); ?

[Reply #6 on:]

Or,
writeTo(DEVICE, 0x16, B00011000);
Or
writeTo(DEVICE, 0x16, 0x18);

Makes it a lot more straightforward which bits are being manipulated.
24, you can't tell where the bits are without doing the conversions.

[Reply #7 on:]

You could also write:
writeTo(DEVICE, 0x16, (3<<3));

That is the value 3 shifted 3 bit to the left.

I assume that you know that you will also be writing a 0 to DLPF_CFG when you do this.

[Reply #8 on:]

which is equal to 24 in decimal, so i'd want:
writeTo(DEVICE, 0x16, 24); ?

While technically this is correct it's unreadable, a day later you'll be wondering WTF the value means.

So I would use the B00011000 or (3<<3) options mentioned. Even better define a value to shift by

#define FS_SEL 3

(3<<FS_SEL)

You will notice that is how most AVR direct port manipulation is written. Straight away you know that the FS_SEL field is being set to 3.

If the DLPF_CFG field is <> 0 then

(3<<FS_SEL) | (dlpf_cfg_val << DLPF_CFG)

or

(3<<FS_SEL) | dlpf_cfg_val

because it's bit offset is 0 anyway.
______
Rob

[Reply #9 on:]

Thanks Graynomad, that approach makes a lot of sense I'll be using that way in new code I write.

[Reply #10 on:]

If you are changing different registers then you might want to make a function to read the register and only change the bits you want changed. Perhaps you will need args for what to & and what to | with the original data before writing back.