失去const挥发性限定词

Question

我有一个编译错误:

error C3848: expression having type 'const unicode::endian_swap<T>'
             would lose some const-volatile qualifiers in order to call
             'unsigned long unicode::endian_swap<T>::operator ()(T &)'

这个错误的描述,在这里,并没有真正解释发生了什么.

我无法在较小的样本中重现错误,但我可以显示我班级的基本布局.

template < typename T >
struct endian_swap
{
    endian_swap ( void ) {}
    T operator () ( T& _val ) const { return _val >> 8 | _val << 8; }
};

template < typename T >
struct test
{
    endian_swap< T > _swap;

    virtual void do_it ( ) const
    {
        unsigned short n = 0x1234;
        unsigned short * _dest = &n;

        *_dest++ = _swap( n ); // <-- Error is here
    }
};

添加endian_swap为成员后弹出错误.实际的类派生自std::codecvt并安装到std::locale.

有人可以比上面的网站更好地解释错误.

编辑:实际代码:

template < typename T, size_t N = sizeof( T ) > struct endian_swap
{
    endian_swap ( void ) {};
    T operator () ( const T _val ) const { return _val };
};
template < typename T > struct endian_swap< T, 2 >
{
    endian_swap ( void ) {}
    T operator () ( const T _val ) const { return _val >> 8 | _val << 8; }
};
template < typename T > struct endian_swap< T, 4 >
{
    endian_swap ( void ) {};
    T operator () ( const T _val ) const { return (_val >> 24) | ((_val & 0x00ff0000) >> 8) | ((_val & 0x0000ff00) << 8) | (_val << 24) };
};

更新:找到它!仔细查看上面的模板,看看你是否也能看到它.

Answer 1

你是如何实例化模板的？您endian_swap使用类型的实例化 T,并传递它unsigned short.除非T是unsigned short,否则您将需要转换,并且转换的结果是临时的,不能绑定到非const引用.

你没有提供SSCCE,所以很难说.不过,在提问的错误消息指的是一个电话unsigned long unicode::endian_swap<T>::operator ()(T &);要么 endian_swap被实例化unsigned long(在这种情况下,通过一个unsigned short需要临时),或者您发布的代码是不是由此引发错误的代码.