[saagarjha]

The problem is the code unconditionally dereferences the pointer, which would be UB if it was a null pointer. This means it is legal to optimize out any code paths that rely on this, even if they occur earlier in program order.

[pwdisswordfishy]

But if the assertion fails, the program is aborted before the pointer would have been dereferenced, making it not UB. This explanation is bogus.

[saagarjha]

Only if the assert is active. It basically means that the code is invalid when NDEBUG is set.

[quuxplusone]

When NDEBUG is set, there is no test, no assertion, at all. So yes, this code has UB if you set NDEBUG and then pass it a null pointer — but that's obvious. The code does exactly what it looks like it does; there's no tricks or time travel hiding here.

[teo_zero]

> it is legal to optimize out any code paths that rely on this, even if they occur earlier in program order.

I don't think this is true. The compiler cannot remove or reorder instructions that have a visible effect.

  if (p == 0)
    printf("Ready?\n");
  *p++;

The printf() can't be omitted.

[aw1621107]

> The compiler cannot remove or reorder instructions that have a visible effect.

You might be surprised! When it comes to UB compilers can and do reorder/eliminate instructions with side effects, resulting in "time travel" [0].

IIRC the upcoming version of the C standard bans this behavior, but the C++ standard still allows it (for now, at least).

[0]: https://devblogs.microsoft.com/oldnewthing/20140627-00/?p=63...

[saagarjha]

No, this is explicitly legal. Most compilers will shy away from it these days since it made a lot of people upset, but it's definitely allowed.

[aw1621107]

> The problem is the code unconditionally dereferences the pointer, which would be UB if it was a null pointer.

Only when NDEBUG is defined, right?

[saagarjha]

No, the code that does this is always active

[gottheUIblues]

Right so strictly speaking C++ could do anything here when passed a null pointer, because even though assert terminates the program, the C++ compiler cannot see that, and there is then undefined behaviour in that case

[aw1621107]

> because even though assert terminates the program, the C++ compiler cannot see that

I think it should be able to. I'm pretty sure assert is defined to call abort when triggered and abort is tagged with [[noreturn]], so the compiler knows control flow isn't coming back.

[aw1621107]

Shouldn't control flow diverge if the assert is triggered when NDEBUG is not defined? Pretty sure assert is defined to call abort when triggered and that is tagged [[noreturn]].

[saagarjha]

Sorry, yes, I misread you