是否可以防止对象的堆栈分配,并且只允许使用"new"实例化它？

Question

是否有可能阻止对象的堆栈分配,并且只允许在堆上使用"new"进行隔离？

Answer 1

一种方法是将构造函数设置为私有,并且只允许通过返回指针的静态方法进行构造.例如:

class Foo
{
public:
    ~Foo();
    static Foo* createFoo()
    {
        return new Foo();
    }
private:
    Foo();
    Foo(const Foo&);
    Foo& operator=(const Foo&);
};

Answer 2

在C++ 11的情况下

class Foo
{
  public:
    ~Foo();
    static Foo* createFoo()
    {
        return new Foo();
    }

    Foo(const Foo &) = delete; // if needed, put as private
    Foo & operator=(const Foo &) = delete; // if needed, put as private
    Foo(Foo &&) = delete; // if needed, put as private
    Foo & operator=(Foo &&) = delete; // if needed, put as private

  private:
    Foo();
};

Answer 3

您可以创建构造函数private,然后提供public静态工厂方法来创建对象.

Answer 4

以下允许公共构造函数,并通过在运行时抛出来停止堆栈分配.注意thread_local是C++ 11关键字.

class NoStackBase {
    static thread_local bool _heap;
protected:
    NoStackBase() {
        bool _stack = _heap;
        _heap = false;
        if (_stack)
            throw std::logic_error("heap allocations only");
    }
public:
    void* operator new(size_t size) throw (std::bad_alloc) { 
        _heap = true;
        return ::operator new(size);
    }
    void* operator new(size_t size, const std::nothrow_t& nothrow_value) throw () {
        _heap = true;
        return ::operator new(size, nothrow_value);
    }
    void* operator new(size_t size, void* ptr) throw () {
        _heap = true;
        return ::operator new(size, ptr);
    }
    void* operator new[](size_t size) throw (std::bad_alloc) {
        _heap = true;
        return ::operator new[](size);
    }
    void* operator new[](size_t size, const std::nothrow_t& nothrow_value) throw () {
        _heap = true;
        return ::operator new[](size, nothrow_value);
    }
    void* operator new[](size_t size, void* ptr) throw () {
        _heap = true;
        return ::operator new[](size, ptr);
    }
};

bool thread_local NoStackBase::_heap = false;