handling operations with macros in C

Question

I'm new to dealing with Macros and I've just stumbled into an exercise that I can't figure it out. Could someone explain to me what's happening here? If I compile i can see what the output is but i can't get it there myself. Thank you in advance!

#define M(varname, index) ( ( (unsigned char*) & varname )[index] ) 

int main(void) {
int a = 0x12345678; 
printf( "%x %x\n", M(a,0), M(a,3) );
printf( "%x %x\n", M(a,1), M(a,2) );
}

It is text substitution: replace the arguments where they occur in the macro definition, with the parameters passed. — Weather Vane, Jun 25 '16 at 14:50
Feed the program through the pre-processor `cpp` to see what the replacement is. After that, it's pointer arithmetic ... — Thomas B Preusser, Jun 25 '16 at 14:50
I can explain "what's happening here" but to get better answer can you please tell us what you expected to see as output and what you actually see? — mvidelgauz, Jun 25 '16 at 14:51

score 6 · Answer 1 · edited May 23 '17 at 11:52

Each Macro usage M(x,y) is replaced with ( (unsigned char*) & x )[y]

So your code looks like this after preprocessing:

int main(void) {
    int a = 0x12345678; 
    printf( "%x %x\n", ( (unsigned char*) & a )[0], ( (unsigned char*) & a )[3] );
    printf( "%x %x\n", ( (unsigned char*) & a )[1], ( (unsigned char*) & a )[2] );
}

Like Thomas B Preusser added in the OP Question comments, most C compiler suites allow to get the pre-processed code with certain compiler flags or tools like f.e. with GCC as mentioned here.

score 3 · Answer 2 · answered Jun 25 '16 at 15:07

Macros work by replacing one thing with another before attempting to compile the code (preprocessing). So in the example you gave:

#define M(varname, index) ( ( (unsigned char*) & varname )[index] )

Each time a M(varname, index) occurs, it will be replaced by ( (unsigned char*) & varname )[index]

In terms of what the code is doing, as an example (reformatted slightly for readability):

printf("%x %x\n", ((unsigned char*) &a)[0], ((unsigned char*) &a)[3]);

This:

takes the address of variable a
casts the address to an unsigned char*
gets the 0th element in this pointer/array
formats it into the string, replacing the first %x, as hex (%x is for hex)

It repeats this for the 3rd element with the second %x

handling operations with macros in C

2 Answers2