Why is the ternary operator used to define 1 and 0 in a macro?

Viktor S picture Viktor S · Mar 31, 2017 · Viewed 9.4k times · Source

I'm using an SDK for an embedded project. In this source code I found some code which at least I found peculiar. In many places in the SDK there is source code in this format:

#define ATCI_IS_LOWER( alpha_char )  ( ( (alpha_char >= ATCI_char_a) && (alpha_char <= ATCI_char_z) ) ? 1 : 0 )

#define ATCI_IS_UPPER( alpha_char )  ( ( (alpha_char >= ATCI_CHAR_A) && (alpha_char <= ATCI_CHAR_Z) ) ? 1 : 0 )

Does the use of the ternary operator here make any difference?

Isn't

#define FOO (1 > 0)

the same as

#define BAR ( (1 > 0) ? 1 : 0)

?

I tried evaluating it by using

printf("%d", FOO == BAR);

and get the result 1, so it seems that they are equal. Is there a reason to write the code like they did?

Answer

Bathsheba picture Bathsheba · Mar 31, 2017

You are correct, in C it is tautologous. Both your particular ternary conditional and (1 > 0) are of type int.

But it would matter in C++ though, in some curious corner cases (e.g. as parameters to overloaded functions), since your ternary conditional expression is of type int, whereas (1 > 0) is of type bool.

My guess is that the author has put some thought into this, with an eye to preserving C++ compatibility.