Search code examples
c++11atomicvolatilestdatomic

Natural alignment + volatile = atomic in C++11?

1) Is the following declaration of a naturally aligned pointer:

alignas(sizeof(void *)) volatile void * p;

equivalent to 

std::atomic<void *>

in C++11?

2) Saying more exactly, is it correct to assume that this type of pointer will work in the same way as std::atomic in C++11?

Solution

  • No, volatile does not guarantee that the location will be written or read atomically, just the the compiler can't optimise out multiple reads and writes.

    On certain architectures, the processor will atomically read or write if aligned correctly, but that is not universal or even guaranteed through a family of processors. Where it can, the internal implementation of atomic will take advantage of architectural features and atomic instruction modifiers, so why not use atomic, if you mean atomic?