I am programming a set of macros able to recognize if a char is in a given range.
The macro is supposed to be used with unsignad chars and returns a boolean

My question: is this set of macros

#define is_range1(x)         (x <    4)
#define is_range2(x)        ((x <    8) && (x >=    4))
#define is_range3(x)        ((x <   16) && (x >=    8))
#define is_range4(x)        ((x <   32) && (x >=   16))
#define is_range5(x)        ((x <   64) && (x >=   32))
#define is_range6(x)        ((x <  128) && (x >=   64))
#define is_range7(x)         (x >  127)

equivalent to this other set of macros

#define is_range1(x)        (x < 4)
#define is_range2(x)        ((x & 4) == 4)
#define is_range3(x)        ((x & 8) == 8)
#define is_range4(x)        ((x & 16) == 16)
#define is_range5(x)        ((x & 32) == 32)
#define is_range6(x)        ((x & 64) == 64)
#define is_range7(x)        (x > 127) 

? And, if not why ?
IMHO the should be completely equivalent.

Actually they are not as the results of the C program
with the second set is completely wrong !!

Let's say, X=96, or $01100000.
Then, your first setup will work: is_range5(x) will return false, because 96 isn't in the domain. is_range6(x) will return true, because 96 is in the domain.
Your second setup will fail: is_range5(x) will return true, as $01100000 AND $00100000 == $00100000. Also, is_range6(x) will return true, as $01100000 AND $01000000 == $01000000.

Those is_range(y)(x) blocks are not equivalent. Only equivalents are is_range1(x) and is_range7(x).

This test is equivalent with the first one, but I'm not sure if it saves you anything:


#define is_range1(x) (x < 4)
#define is_range2(x) ((x & ~3) == 4)
#define is_range3(x) ((x & ~7) == 8)
#define is_range4(x) ((x & ~15) == 16)
#define is_range5(x) ((x & ~31) == 32)
#define is_range6(x) ((x & ~63) == 64)
#define is_range7(x) (x > 127)

Note that the compiler will calculate the ~'ed values for you so its not done runtime.

I see my error!
thanks

it woks!

PS
in my program, the second method saves about 100 LM instructions

It is good practice to protect the arguments of macros with parentheses.
#define is_range4(x) ((x & ~15) == 16)
will not give the right results for is_range4(z+3) because it gets replaced with ((z+3 & ~15) == 16), which will be simplified by the compiler too ((z+3) == 16).
The proper way to define the macros is like this:
#define is_range4(x) (((x) & ~15) == 16)
Macros look just like functions in your code, but they are substitutions in the text handled by the preprocessor.

thnx