有没有理由不在一个解析为gcc中的__builtin_unreachable（）的宏中包装assert（）？

Question

Context: In this answer , I learned that gcc's __builtin_unreachable() can have some surprisingly impactful performance implications, as it seems that the following:

if(condition) __builtin_unreachable();

is being entirely stripped, and used as an optimization hint as long as condition can be guaranteed to not have any side effect.

So my immediate reaction to this is that I should create the following macro, and use it absolutely everywhere I would normally use assert() , since side-effect inducing code inside an assert() would be a major bug in the first place:

// TODO: add handling of other compilers as appropriate.
#if defined(__GNUC__) && defined(NDEBUG)
  #define my_assert(condition) \
    if(!(condition)) __builtin_unreachable()
#else 
  #define my_assert(condition) assert(condition)
#endif

From a standards perspective, this would create a split in functionality between normal and NDEBUG builds, which you could make an argument excludes this macro from being the standard behavior of assert() . However, since my code would be functionally dead in the water in the case of assertion failures regardless, it's fully equivalent from a behavioral standpoint.

So my question is: Can anyone think of a reason not to do this (appart from asserts that involve a lot of indirections)?

Before you ask, yes, I've checked that gcc's behavior is to void cast the assertion away in NDEBUG builds.

Answer 1

Yes, there is a reason to not use it. Some people use the following defensive code practice that combines assert and exception ( assert(x>0); if (!(x<0)) throw std::logic_error("..") ) - see this answer:

Test Cases AND assertion statements

Your macro silently breaks the exception-part for release builds.