What is the Macro?
If you write as “#define” in your code, that is “Macro” .

Macro will handle by the preprocessor.
You can think Macro as a method to replace the String which appears in you code..

pros:
– run faster than writing as function

cons:
– occupy memories

concepts:
– there is no the concept of pointer in macro.
just “replacing the string”

An question:
What is the output of the following code?

#define macro_to_func(pp) ({printf("in Macro ...\n"); minus ;})

void do_something(int *a, int *b){
    printf("do_something\n");
    int c;
    c = *a + *b;
    printf("do_something %d \n", c);
}

int minus(int *a, int *b){
    printf("minus\n");
    int c;
    c = *a - *b +1;
    printf("minus a = %d \n", *a);
    printf("minus b = %d \n", *b);
    printf("minus %d \n", c);
    return c;
}

int main()
{
    int (*do_something)(int* c, int* d);
    //printf("fptr 1 do_something %p\n", &do_something);

    int m = -1;
    int n = 7;

    printf("do_something = macro_to_func\n");
    do_something = macro_to_func(pp);
    //printf("fptr minus %p\n", &minus);
    //printf("fptr 2 do_something %p\n", &do_something);
    if(do_something){
        printf("in IF!!\n");
        do_something(&m, &n); --------------------------->???
    }
    return 0;
}

Answer:

do_something = macro_to_func
in Macro …
in IF!!
minus
minus a = -1
minus b = 7
minus -7

Explanation:
“int (*do_something)” is a local function pointer.
In the macro, the macro_to_func will be replaced as “minus”, but do_something !!!