Question

How would I define multiple values of the same type in an array using #define ?

For instance, I would like

#define DIGIT 0x30 | 0x31 | 0x32 | 0x33 | 0x34 | 0x35 | 0x36 | 0x37 | 0x38 | 0x39
#define QUOTE 0x22 | 0x27

ETC...

Answer 1

How would I define multiple values of the same type in an array using #define ?
For instance, I would like

#define DIGIT 0x30 | 0x31 | 0x32 | 0x33 | 0x34 | 0x35 | 0x36 | 0x37 | 0x38 | 0x39
#define QUOTE 0x22 | 0x27

Well, the term array in C and C++ refers to a number of variables of the same type, all side by side in memory. If you really wanted an array, then you can use:

static const char digits[] = {
    0x30, 0x31, 0x32, 0x33, 0x34, 0x35, 0x36, 0x37, 0x38, 0x39
};

Of course, there’s nothing to stop you putting part of this in a preprocessor macro, but no obvious point either, and macros are best avoided as conflicting and unintended substitutions aren’t always handled well by the compiler:

#define DIGITS 0x30, 0x31, 0x32, 0x33, 0x34, 0x35, 0x36, 0x37, 0x38, 0x39
static const char digits[] = { DIGITS };

If your intension it to check whether a particular character is one of the listed characters, then you can do it in many ways:

if (isdigit(c)) ...    // use a library function

static const char digits[] = {
    0x30, 0x31, 0x32, 0x33, 0x34, 0x35, 0x36, 0x37, 0x38, 0x39, 0
};
if (strchr(digits, c)) ...  // use another library function (probably slower)

static const char digits[] = "0123456789";  // exactly the same as above!
if (strchr(digits, c)) ...

if (c == 0x30 || c == 0x31 || c == 0x32 ...) ...    // painfully verbose!

if (c == '0' || c == '1' || c == '2' ...) ... // verbose but self-documenting

if (c >= '0' && c <= '9')  // as per caf's comment, C requires the
                           // character set to have contiguous numbers