Question

在c ++中，对分配有delete的对象调用new会调用该类的析构函数，并释放内存。如果不是显式调用其析构函数然后释放内存，而不是删除对象，会有什么区别？

请考虑以下示例。

#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;
}

使用g++ 7.3.0和clang 6.0.1以及-Wextra -Wall -pedantic编译代码不会产生任何警告。该程序的输出与预期的一样

Destructor, i was 3

使用valgrind / memcheck运行程序会产生不匹配的释放/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，上面代码中处理对象p的标准方法是调用delete。我的问题更正式：

delete完全等同于对析构函数的调用和 free？标准是否指定了这一点，还是我遇到了UB 上面的代码？

Answer 1

虽然合法地调用析构函数是合法的，但在这种情况下，您极少会使用析构函数（例如，在较大缓冲区中构造和破坏对象的情况下，例如std :: vector）。

请注意，您应始终将分配器与适当的内存释放，malloc / free，new / delete等进行匹配。虽然操作符new（）通常在幕后依赖于malloc，但并不需要这样做那样的话，您的不匹配将产生不确定的结果。

删除与显式调用析构函数

1 个答案: