delete与析构函数的显式调用[重复]

Question

在c语言中，对分配了new的对象调用delete，调用类的析构函数，并释放内存。如果不删除对象，而是显式调用其析构函数，然后释放内存，会有什么不同吗？

例如，考虑以下示例。

#include <iostream>

struct A {
  int i;
  ~A() { std::cout << "Destructor, i was " << i << std::endl; }
};

int main()
{
  A* p = new A{3};
  p->~A();
  free(p);

  return 0;
}

用7.3.0和6.0.1以及Wextra-Wall-pedantic编译代码不会产生任何警告。正如预期的那样，程序的输出是

Destructor, i was 3

使用valgrind/memcheck运行程序会出现不匹配的释放/删除错误：

==27743== Memcheck, a memory error detector
==27743== Copyright (C) 2002-2017, and GNU GPL'd, by Julian Seward et al.
==27743== Using Valgrind-3.14.0 and LibVEX; rerun with -h for copyright info
==27743== Command: ./destructor
==27743== 
Destructor, i was 3
==27743== Mismatched free() / delete / delete []
==27743==    at 0x4C3033B: free (vg_replace_malloc.c:530)
==27743==    by 0x4009FC: main (in /tmp/destructor)
==27743==  Address 0x5bbac80 is 0 bytes inside a block of size 4 alloc'd
==27743==    at 0x4C2F77F: operator new(unsigned long) (vg_replace_malloc.c:334)
==27743==    by 0x4009DA: main (in /tmp/destructor)
==27743== 
==27743== 
==27743== HEAP SUMMARY:
==27743==     in use at exit: 0 bytes in 0 blocks
==27743==   total heap usage: 3 allocs, 3 frees, 73,732 bytes allocated
==27743== 
==27743== All heap blocks were freed -- no leaks are possible
==27743== 
==27743== For counts of detected and suppressed errors, rerun with: -v
==27743== ERROR SUMMARY: 1 errors from 1 contexts (suppressed: 0 from 0)

但是，没有内存泄漏。

我当然知道，在上面的代码中处理对象的标准方法是调用delete。我的问题比较正式：

• delete是否完全等同于调用析构函数和free？标准是否规定了这一点，或者我用上面的代码遇到了UB

Answer 1

虽然显式调用析构函数是合法的，但您希望这样做的情况非常罕见（例如，在较大缓冲区内构造和销毁对象的情况下）。

请注意，您应该始终将分配器与适当的内存释放、malloc/free、new/delete等匹配。虽然运算符new（）通常在后台依赖malloc，但并不要求它这样做，在这种情况下，您的不匹配将产生未定义的结果。