Explanation for assert macro

Question

I couldn't comment on the answer itself, so: about Using comma to prevent the need for brace pair

 #define MY_ASSERT(expr) ((expr) || (debugbreak(), 0))

Here debugbreak() returns void, but we still wish to have 0 as an rvalue.

How does (debugbreak(), 0) work to return 0? I understand that the return value of debugbreak() is discarded and 0 is returned, but debugbreak generates an exception, so how can anything be evaluated afterward? I suppose my question can be generalised to any similar binary operator where the first part being evaluated exits the program.

Answer 1

It's a type system hack.

#define MY_ASSERT(expr) ((expr) || (debugbreak(), 0))
//   note the or operator here  ^

The || operator takes two bool-typed (or convertible) expressions. debugbreak() is void-typed. To make it bool, use the following rule:

(FOO, BAR)
//    ^  determines the type of the entire comma expression

This is the same as {FOO; BAR} except that a block (in braces) has no type.

Answer 2

Nothing will be evaluated if the assert fires, but both expressions must have the right return type, otherwise this macro will just break compilation.