Question

let me explain my confusion with sample code

this is our macro

#define rsAssert(v)   if(!(v)) printf("rsAssert failed: %s, in %s at %i" #v, __FILE__, __LINE__); 

case 1:

int main(void)
{
  rsAssert(0);
  return 0;
}

this case compiles succesfully

case 2

int main(void)
{
  rsAssert(0)  // note here ;  is not present
  return 0;
}

this also compiles succsfully
Question 1:

it means whether you write rsAssert(0) or rsAssert(0); no difference between them?

---

then

case 3

int main()
{
  if(1)
    rsAssert(0);
  else 
    printf("mr.32");
  return 0;
}

here rsassert(0); is not going to compile [see http://ideone.com/7dFv1%5D but without ; rsasser(0) works fine [see http://ideone.com/8fehl%5D..

I am not getting what’s going on with macro expansion here…

Answer 1

The reason it is not working in case 3 is because this is what is actually getting compiled:

int main()
{
if(1)
  if(!(v)) printf("rsAssert failed: %s, in %s at %i", #v, __FILE__, __LINE__);;
else 
  printf("mr.32");
return 0;
}

Note the extra semicolon at the end of the first printf from the macro.

Normally, with an if statement, it’s followed by a single statement, or a block with { }.

But because of the extra ;, you have this:

if (...) 
    printf(...); ;
else
    printf(...);
return 0;

That extra semicolon is an empty statement, but is still a statement nonetheless.

You should remove the ; from the macro. Then things would make more sense.