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0 pointsagent3274y ago0 comments

>With "volatile variables" you can use compound assignment operators on the variable. What does that even mean? Nothing.

It means exactly the same thing as on a normal variable, and it boggles the mind that people somehow not understand that. Given 'volatile int i', 'i++' means the exact same thing as 'i = i + 1'. Does that also not make any sense to you? If it does, can you explain why you believe they are different?

Volatile member functions and parameters make no sense, but volatile member variables most certainly do. And there is considerable pushback in the C++ community because this is a significant loss of compatibility with various C-headers used frequently in embedded applications. I wouldn't be surprised if the deprecated features will be reinstated in the language in the end.

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tialaramex4y ago

> It means exactly the same thing as on a normal variable, and it boggles the mind that people somehow not understand that.

If you just want normal variables, write a normal variable, J F. Bastien's change doesn't have any consequences for your normal variables.

> Given 'volatile int i', 'i++' means the exact same thing as 'i = i + 1'

No. That's the post-increment operator so i++ has the same value as i did before the increment, whereas i = i + 1 has the value of i after the increment.

And you might say "Oh, but I never use those values". Good. J.F. Bastien's change forbids that for volatiles too, we're on the same page.

What you apparently want is to do two volatile operations, one read and one write, sandwiching an arbitrary integer operation. You can do that explicitly of course, Bastien's change doesn't alter that - but when you write it out, doubt appears on the horizon. If these are two operations, can't I race with somebody else such that they modify this thing before I write my modified version back and then I overwrite their changes?

And that doubt should be there. But what we know from asking people is that the compound operators falsely reassure them. That's why they were deprecated and why it's sad that there's enthusiasm to bring them back.

gpderetta4y ago

Yes, it will likely be reverted [1] for bitwise ops, while keeping += and ++ deprecated (to save face someone would say).

[1] http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2021/p232...

tialaramex4y ago

It's true that the relevant committee is now likely to propose this be changed in C++ 23 (or before as an errata)

But the face saving is on the part of embedded C++ proponents who'd rather never fix the C++ language than risk that the mundane C code cheap vendors actually ship can't be counted as "C++ support" because it's no longer valid C++.

That's the core of the argument in P2327. Not that this isn't a bad idea (it's obviously a bad idea) or that we shouldn't do better (alternatives to C++ already do better) but that this is the status quo and C++ can't expect to improve upon that. A sad state of affairs.

P2327's dismissal of the problem caused by UART1−>UCSR0B |= (1<<UCSZ01 ); comes down to the usual Real Programmer stance, surely every single embedded programmer will arrange "by construction" that the problem can't happen. No actual examples were examined to see if embedded developers reliably do that, which seems unlikely - we're just invited to imagine that they're all inerrant.

vlovich1234y ago

Not sure what will actually happen, but you could easily allow volatile variables with if they have extern “C” linkage for compatibility with C headers, while deprecating it elsewhere.

agent327OP4y ago

That's not really what extern "C" does though. It doesn't change the language rules or the parsing or anything, it only changes how symbols are represented in the object file. Extern "C" means they are mangled using C rules (that mostly involves adding an underscore); otherwise it's gonna be C++ rules (and loads more information needs to be encoded).

Some people argue for a similar mechanism, something like 'language "C++20" { .. }', that would allow a program to opt in to changes that would otherwise be breaking changes; mostly new keywords. However, changing actual language rules in such a block would be tricky, to say the least.

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tialaramex4y ago

> It means exactly the same thing as on a normal variable, and it boggles the mind that people somehow not understand that.

If you just want normal variables, write a normal variable, J F. Bastien's change doesn't have any consequences for your normal variables.

> Given 'volatile int i', 'i++' means the exact same thing as 'i = i + 1'

No. That's the post-increment operator so i++ has the same value as i did before the increment, whereas i = i + 1 has the value of i after the increment.

And you might say "Oh, but I never use those values". Good. J.F. Bastien's change forbids that for volatiles too, we're on the same page.

gpderetta4y ago

Yes, it will likely be reverted [1] for bitwise ops, while keeping += and ++ deprecated (to save face someone would say).

[1] http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2021/p232...

tialaramex4y ago

It's true that the relevant committee is now likely to propose this be changed in C++ 23 (or before as an errata)

vlovich1234y ago

Not sure what will actually happen, but you could easily allow volatile variables with if they have extern “C” linkage for compatibility with C headers, while deprecating it elsewhere.

agent327OP4y ago

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