Confusion in the use of byte variable

Question

A byte stores an 8-bit unsigned number, from 0 to 255.

I can understand the following line:

byte b = B10010;  // "B" is the binary formatter (B10010 = 18 decimal)

But I also saw a use such as:

volatile byte state = LOW;

What does the last line mean here? Is that special for Arduino? I mean LOW is not a 8 bit unsigned number. LOW is not a number here.

score 4 · Answer 1 · answered Jan 24 '17 at 13:59

There is probably somewhere else, like this:

#define LOW 0

Which means the pre-processor turns your line from this:

volatile byte state = LOW;

into this:

volatile byte state = 0;

There is nothing arduino-specific here, just the use of standard C/C++ features.

score 2 · Answer 2 · answered Jan 24 '17 at 14:25

2

It is common practise in C/C++ to define certain much-used constants at the beginning of the code.

#define true 1

this means that every time you write true in your code, the compiler will see it as 1.

The arduino IDE comes with a few constants predefined:

answered Jan 24 '17 at 14:25

Janw

in C++ true and false are already keywords, why would one try to #define them? – jfpoilpret Jan 25 '17 at 07:28
@jfpoilpret Perhaps if one is not using C++ ... – Marcel Besixdouze Mar 21 '21 at 19:34

score 0 · Answer 3 · answered Jan 24 '17 at 13:59

In the same way that you can do

int blarb = 12;
#define wibble 73

you can do

int FOO = 73;
#define BINKY 99

or even

int LOW = 0;
#define HIGH 1

So in the line

volatile byte state = LOW;

someone has created some sort of numerical constant with the name LOW.

You can read all about various constants provided for you here.

3 Answers3