> The buffer[x].available = 0 will never run while buffer[x].available is equal to 0 overwriting or causing a data race when setting buffer[x].available to 1.

In particular, because loads and stores of the same variable cannot be reordered out of program order. Once your algorithm involves other variables, you would (likely) need to be a little careful about loading/storing with acquire/release semantics to prevent reordering other accesses relative to this protocol.

> Remember to use compiler memory barrier

I would highly recommend using the language atomic types (and barriers if truly needed) instead of gcc inline assembly syntax.