Destructor implicitly calls the Finalize method, they are technically the same. Dispose is available with objects that implement the IDisposable interface. You may see : Destructors C# – MSDN The destructor implicitly calls Finalize on the base class of the object. Example from the same link: class Car { ~Car() // destructor { // cleanup … Read more
If you call exit, the destructor will not be called. From the C++ standard (ยง3.6.1/4): Calling the function void exit(int); declared in <cstdlib> (18.3) terminates the program without leaving the current block and hence without destroying any objects with automatic storage duration (12.4). If exit is called to end a program during the destruction of … Read more
It does for C# (see code below) but not for C++. using System; class Test { Test() { throw new Exception(); } ~Test() { Console.WriteLine(“Finalized”); } static void Main() { try { new Test(); } catch {} GC.Collect(); GC.WaitForPendingFinalizers(); } } This prints “Finalized”
A temporary variable lives until the end of the full expression it was created in. Yours ends at the semicolon. This is in 12.2/3: Temporary objects are destroyed as the last step in evaluating the full-expression (1.9) that (lexically) contains the point where they were created. Your behavior is guaranteed. There are two conditions that, … Read more
Within each category of storage classes (except dynamically allocated objects), objects are destructed in the reverse order of construction.
You’re looking for unset(). But take into account that you can’t explicitly destroy an object. It will stay there, however if you unset the object and your script pushes PHP to the memory limits the objects not needed will be garbage collected. I would go with unset() (as opposed to setting it to null) as … Read more
if a constructor throws an exception, what destructors are run? Destructors of all the objects completely created in that scope. Does it make any difference if the exception is during the initialization list or the body? All completed objects will be destructed. If constructor was never completely called object was never constructed and hence cannot … Read more
No, by default, most signals cause an immediate, abnormal exit of your program. However, you can easily change the default behavior for most signals. This code shows how to make a signal exit your program normally, including calling all the usual destructors: #include <iostream> #include <signal.h> #include <unistd.h> #include <cstring> #include <atomic> std::atomic<bool> quit(false); // … Read more
Yes, automatic variables will be destroyed at the end of the enclosing code block. But keep reading. Your question title asks if a destructor will be called when the variable goes out of scope. Presumably what you meant to ask was: will Foo’s destructor be called at the end of main()? Given the code you … Read more
It will do the same thing (nothing, in essence). But it’s not the same as if you didn’t write it. Because writing the destructor will require a working base-class destructor. If the base class destructor is private or if there is any other reason it can’t be invoked, then your program is faulty. Consider this … Read more