C++ 0x仍然显式地用全局运算符新分配吗?

Wikipedia

struct NonNewable {
    void *operator new(std::size_t) = delete;
};

此类型的对象只能作为堆栈对象或其他类型的成员进行分配。如果没有不可移植的技巧,它就不能直接堆分配。(由于placement new是在用户分配的内存上调用构造函数的唯一方法,并且如上所述禁止这种使用,因此无法正确构造对象。)

删除运算符new类似于在当前C++中使其私有,但不明确地使用全局运算符new,从而避免了类特定查找,仍然有效C++ +0x?

NonNewable *p = ::new NonNewable();
// neither non-portable nor trickery, though perhaps not widely known

works fine :

struct NonNewable {
private:
  void *operator new(std::size_t);  // not defined
};

int main() {
  // ignore the leaks, it's just an example

  void *mem = operator new(sizeof(NonNewable));
  NonNewable *p = ::new(mem) NonNewable();

  p = ::new NonNewable();

  return 0;
}