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If a variable has block scope in C++ (i.e. it is a local variable in a function) then its destructor is guaranteed to be called when the block is finished, rega

If a variable has block scope in C++ (i.e. it is a local variable in a function) then its destructor is guaranteed to be called when the block is finished, regardless of whether that is due to a `return` statement or an exception being thrown (or a `break` or `continue`). In what sense do you disagree? If you allocate an object on the heap with `new` then its destructor isn't called automatically unless you make it so through some other mechanism, but GP comment clearly want claiming that. There are some situations where objects with block scope do not have their destructor called e.g. `_exit()` called, segfault, power cable pulled out. But in that sense nothing is guaranteed.

DanWaterworth 8y ago
If you are consuming an API that provides an object with a destructor, you are correct, you can determine when destructors will be called. The issue is when you produce an API that contains objects with destructors. Since you are handing these entities off to unknown code, you cannot ensure that they will be dropped. This was a problem in scoped threads in Rust.
- siscia 8y ago
  Can you please dig deeper, that I am not sure I follow. In which case in rust you cannot be sure that "the drop" will be called?
  - richardwhiuk 8y ago
    A Rc cycle causing a leak. See the very excellent http://cglab.ca/~abeinges/blah/everyone-poops/ http://cglab.ca/~abeinges/blah/everyone-poops/
  - dbaupp 8y ago
    If there's a cycle of strong references with Rc or Arc (or shared_ptr in C++), those objects still never get dropped/have their destructors called.
    - quietbritishjim 8y ago
      I think that falls into the category I mentioned in the third paragraph of my comment: a serious pre-existing bug with other consequences will potentially cause the guarantee to be violated. A similar effect would happen if you had a double free that sometimes caused a crash, which is a similar level of programming mistake to creating a cyclic reference. To me it sits outside of a reasonable definition of "guaranteed".
      - dbaupp 8y ago
        No, typically, a reference cycle is fine. It results in valid memory that never gets read again, which is unfortunate but not dangerous, whereas double-frees can result in memory corruption. http://huonw.github.io/blog/2016/04/memory-leaks-are-memory-safe/ http://huonw.github.io/blog/2016/04/memory-leaks-are-memory-...
    - zaphar 8y ago
      Rc's drop will be called. But whether the exposed object's drop will be called is dependent on the reference count. But Rc would not work if the drop was not guaranteed to be called.
      - steveklabnik 8y ago
        If you have a reference cycle, the two Rcs will keep each other alive, and their Drops will not be called.
      - dbaupp 8y ago
        I was a little unclear but that is of course what I meant: talking about the underlying shared data because the pointers themselves don't have particularly interesting destruction behaviour. (Although the sibling is also correct that not all Rc/Arc/shared_ptr handles to the shared data with have their Drop called.)
flurrything 8y ago
> In what sense do you disagree? Not the parent, but it is trivial to write C++ and Rust examples in which destructors of variables with block scope are not called. The std library of both languages do even come with utilities to do this: C++ structs: struct Foo { Foo() { std::cout << "Foo()" << std::endl; } ~Foo() { std::cout << "~Foo()" << std::endl; } }; { std::aligned_storage<sizeof(Foo),alignof(Foo)> foo; new(&foo) Foo; /* destructor never called even though a Foo lives in block scope and its storage is free'd */ } C++ unions: union Foo { Foo() { std::cout << "Foo()" << std::endl; } ~Foo() { std::cout << "~Foo()" << std::endl; } }; { Foo foo(); /* destructor never called */ } Rust: struct Foo; impl Drop for Foo { fn drop(&mut self) { println!("drop!"); } } { let _ = std::mem::ManuallyDrop::<Foo>::new(Foo); /* destructor never called */ } etc. > There are some situations where objects with block scope do not have their destructor called e.g. `_exit()` called, segfault, power cable pulled out. But in that sense nothing is guaranteed. This is pretty much why it is impossible for a programming language to guarantee that destructors will be called. Might seem trivial, but even when you have automatic storage, any of the things you mention can happen, such that destructors won't be reached. In general, C++, Rust, etc. cannot guarantee that destructors will be called, because it is also trivial to make that impossible once you start using the heap (e.g. a `shared_ptr` cycle will never be freed).

DanWaterworth 8y ago

If you are consuming an API that provides an object with a destructor, you are correct, you can determine when destructors will be called. The issue is when you produce an API that contains objects with destructors. Since you are handing these entities off to unknown code, you cannot ensure that they will be dropped. This was a problem in scoped threads in Rust.

siscia 8y ago
Can you please dig deeper, that I am not sure I follow. In which case in rust you cannot be sure that "the drop" will be called?
- richardwhiuk 8y ago
  A Rc cycle causing a leak. See the very excellent http://cglab.ca/~abeinges/blah/everyone-poops/ http://cglab.ca/~abeinges/blah/everyone-poops/
- dbaupp 8y ago
  If there's a cycle of strong references with Rc or Arc (or shared_ptr in C++), those objects still never get dropped/have their destructors called.
  - quietbritishjim 8y ago
    I think that falls into the category I mentioned in the third paragraph of my comment: a serious pre-existing bug with other consequences will potentially cause the guarantee to be violated. A similar effect would happen if you had a double free that sometimes caused a crash, which is a similar level of programming mistake to creating a cyclic reference. To me it sits outside of a reasonable definition of "guaranteed".
    - dbaupp 8y ago
      No, typically, a reference cycle is fine. It results in valid memory that never gets read again, which is unfortunate but not dangerous, whereas double-frees can result in memory corruption. http://huonw.github.io/blog/2016/04/memory-leaks-are-memory-safe/ http://huonw.github.io/blog/2016/04/memory-leaks-are-memory-...
  - zaphar 8y ago
    Rc's drop will be called. But whether the exposed object's drop will be called is dependent on the reference count. But Rc would not work if the drop was not guaranteed to be called.
    - steveklabnik 8y ago
      If you have a reference cycle, the two Rcs will keep each other alive, and their Drops will not be called.
    - dbaupp 8y ago
      I was a little unclear but that is of course what I meant: talking about the underlying shared data because the pointers themselves don't have particularly interesting destruction behaviour. (Although the sibling is also correct that not all Rc/Arc/shared_ptr handles to the shared data with have their Drop called.)

flurrything 8y ago

> In what sense do you disagree? Not the parent, but it is trivial to write C++ and Rust examples in which destructors of variables with block scope are not called. The std library of both languages do even come with utilities to do this: C++ structs: struct Foo { Foo() { std::cout << "Foo()" << std::endl; } ~Foo() { std::cout << "~Foo()" << std::endl; } }; { std::aligned_storage<sizeof(Foo),alignof(Foo)> foo; new(&foo) Foo; /* destructor never called even though a Foo lives in block scope and its storage is free'd */ } C++ unions: union Foo { Foo() { std::cout << "Foo()" << std::endl; } ~Foo() { std::cout << "~Foo()" << std::endl; } }; { Foo foo(); /* destructor never called */ } Rust: struct Foo; impl Drop for Foo { fn drop(&mut self) { println!("drop!"); } } { let _ = std::mem::ManuallyDrop::<Foo>::new(Foo); /* destructor never called */ } etc. > There are some situations where objects with block scope do not have their destructor called e.g. `_exit()` called, segfault, power cable pulled out. But in that sense nothing is guaranteed. This is pretty much why it is impossible for a programming language to guarantee that destructors will be called. Might seem trivial, but even when you have automatic storage, any of the things you mention can happen, such that destructors won't be reached. In general, C++, Rust, etc. cannot guarantee that destructors will be called, because it is also trivial to make that impossible once you start using the heap (e.g. a `shared_ptr` cycle will never be freed).