class X { /* ... */};
void* raw = malloc(sizeof (X));
X* p = new (raw) X(); // according to the standard,the RHS is a placement-new expression
::operator delete(p); // definitely wrong,per litb's answer
delete p; // legal?  I hope not

也许你们中的一位语言律师可以解释这个标准是禁止这个.

还有一个数组形式：

class X { /* ... */};
void* raw = malloc(sizeof (X));
X* p = new (raw) X[1]; // according to the standard,the RHS is a placement-new expression
::operator delete[](p); // definitely wrong,per litb's answer
delete [] p; // legal?  I hope not

This is the closest question我能找到.

编辑：我只是不认为标准的语言限制参数的函数void :: operator delete(void *)的用法以任何有意义的方式应用于delete-expression中的delete操作数.最好,两者之间的连接是非常脆弱的,并且允许一些表达式作为删除的操作数,这些操作不能传递给void :: operator delete(void *).例如：

struct A
{
  virtual ~A() {}
};

struct B1 : virtual A {};

struct B2 : virtual A {};

struct B3 : virtual A {};

struct D : virtual B1,virtual B2,virtual B3 {};

struct E : virtual B3,virtual D {};

int main( void )
{
  B3* p = new E();
  void* raw = malloc(sizeof (D));
  B3* p2 = new (raw) D();

  ::operator delete(p); // definitely UB
  delete p; // definitely legal

  ::operator delete(p2); // definitely UB
  delete p2; // ???

  return 0;
}

我希望这可以说明一个指针是否可以被传递给void operator delete(void *)与该指针是否可以用作delete的操作数无关.