Confused about C macro expansion and integer arithmetic [duplicate]

Answer 1

Regarding your question about

#define AA (-64)

See Macro definition and expansion in the C preprocessor:

Object-like macros were conventionally used as part of good programming practice to create symbolic names for constants, e.g.

#define PI 3.14159

... instead of hard-coding those numbers throughout one's code. However, both C and C++ provide the const directive, which provides another way to avoid hard-coding constants throughout the code.

Constants defined as macros have no associated type. Use const where possible.

Answer 2

sizeof returns the number of bytes in unsigned format. That's why you need the cast.

See more here.

Answer 3

Answering just one of your sub-questions:

To "define a constant in a macro" (this is a bit sloppy, you're not defining a "constant", merely doing some text-replacement trickery) that is unsigned, you should use the 'u' suffix:

#define UNSIGNED_FORTYTWO 42u

This will insert an unsigned int literal wherever you type UNSIGNED_FORTYTWO.

Likewise, you often see (in <math.h> for instance) suffices used to set the exact floating-point type:

#define FLOAT_PI 3.14f

This inserts a float (i.e. "single precision") floating-point literal wherever you type FLOAT_PI in the code.

Answer 4

Look at this line:

for(d=-1;d <= (TOTAL_ELEMENTS-2);d++)

In the first iteration, you are checking whether

-1 <= (TOTAL_ELEMENTS-2)

The operator size_of returns unsigned value and the check fails (-1 signed = 0xFFFFFFFF unsigned on 32bit machines).

A simple change in the loop fixes the problem:

for(d=0;d <= (TOTAL_ELEMENTS-1);d++)
printf("%d\n",array[d]);

To answer your other questions: C macros are expanded text-wise, there is no notion of types. The C compiler sees your loop as this:

for(d=-1;d <= ((sizeof(array) / sizeof(array[0]))-2);d++)

If you want to define an unsigned constant in a macro, use the usual suffix (u for unsigned, ul for unsigned long).