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How does that interact with using C++ to access hardware features via direct memory, such as UART registers on embedded systems? Is all of that considered UB be
by shadowgovt 2y ago
How does that interact with using C++ to access hardware features via direct memory, such as UART registers on embedded systems? Is all of that considered UB because the compiler can't prove there's anything at the UART "magic" addresses?
- int_19h 2y agoAs far as ISO C++ is concerned, it is. In practice, of course, a useful C++ compiler would let you do it, somehow. But that is firmly in the realm of implementation-specific stuff (keeping in mind that UB can mean anything including being well-defined by a specific implementation.