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> AIUI, there are some hardware architectures where even creating a wild pointer might be undefined behavior, regardless of whether that pointer is subsequently

> AIUI, there are some hardware architectures where even creating a wild pointer might be undefined behavior, regardless of whether that pointer is subsequently dereferenced Would you be able to point me to some references for such hardware? Im not sure how that would work (at least based on my admittedly limited amount of experience). Wouldn’t a pointer look like any other integer right up until it’s used as a memory operand? Or would said architecture have some way to distinguish pointers and a “regular” integer in registers?

Hello71 7y ago
Allegedly, some platforms have pointer trap representations, where a certain pointer can be created, but may not be used in any operations of that type. No modern systems have such trap representations for pointer types, but the C standard inherits their legacy, and, more importantly, C compilers use it as justification for certain types of optimizations. Since it's not a hardware limitation, Rust can perfectly well take the opposite path and say that the compiler may not use it for those optimizations. https://stackoverflow.com/questions/6725809/trap-representation https://stackoverflow.com/questions/6725809/trap-representat...
- zozbot234 7y ago
  > No modern systems have such trap representations for pointer types This may be incidentally true, but "address sanitizer"-like features are becoming more common on modern hardware, and while these do not currently trap on creation/manipulation of a 'wild' pointer (since, strictly speaking, a trap only happens on dereferencing), there's no solid reason to expect this to remain the case in the future.
  - int_19h 7y ago
    I don't see how you could trap creation or manipulation, since those pointers are stored in registers and/or memory, and both are fundamentally untyped. How would the hardware even know that something is a pointer, on any architecture that is popular today?
    - saagarjha 7y ago
      Because you use typed instructions to access them. For example, on ARM with pointer authentication you’ll sign pointers and unsign them right before using them. If you forge a pointer it’ll cause a crash when it’s used because its signature will be incorrect.
      - int_19h 7y ago
        But that would still happen at the point of dereference, no? Or does it allow to tag even operations like moves and adds?
- nicoburns 7y ago
  > C compilers use it as justification for certain types of optimizations I believe that the Rust compiler is free to make it's own choices about what is considered valid, and which optimisations it wants to enable. It doesn't need to follow C's lead here.
  - Hello71 7y ago
    isn't that what I said

Hello71 7y ago

Allegedly, some platforms have pointer trap representations, where a certain pointer can be created, but may not be used in any operations of that type. No modern systems have such trap representations for pointer types, but the C standard inherits their legacy, and, more importantly, C compilers use it as justification for certain types of optimizations. Since it's not a hardware limitation, Rust can perfectly well take the opposite path and say that the compiler may not use it for those optimizations. https://stackoverflow.com/questions/6725809/trap-representation https://stackoverflow.com/questions/6725809/trap-representat...

zozbot234 7y ago
> No modern systems have such trap representations for pointer types This may be incidentally true, but "address sanitizer"-like features are becoming more common on modern hardware, and while these do not currently trap on creation/manipulation of a 'wild' pointer (since, strictly speaking, a trap only happens on dereferencing), there's no solid reason to expect this to remain the case in the future.
- int_19h 7y ago
  I don't see how you could trap creation or manipulation, since those pointers are stored in registers and/or memory, and both are fundamentally untyped. How would the hardware even know that something is a pointer, on any architecture that is popular today?
  - saagarjha 7y ago
    Because you use typed instructions to access them. For example, on ARM with pointer authentication you’ll sign pointers and unsign them right before using them. If you forge a pointer it’ll cause a crash when it’s used because its signature will be incorrect.
    - int_19h 7y ago
      But that would still happen at the point of dereference, no? Or does it allow to tag even operations like moves and adds?
nicoburns 7y ago
> C compilers use it as justification for certain types of optimizations I believe that the Rust compiler is free to make it's own choices about what is considered valid, and which optimisations it wants to enable. It doesn't need to follow C's lead here.
- Hello71 7y ago
  isn't that what I said