我想做一些看起来很简单使用模板,但我不能让它工作。该类正在为AVR处理器实现各种串行IO实现,但问题是一个通用的C++问题。目标是在编译时根据模板参数制作选项,以便用户友好并增加代码重用,并在某些地方由于重用而提高性能。C++模板部分类专业化与功能专业
问题很简单,但我找不到解决方案(如果有的话)。当编译下面使用Visual Studio 2008的代码我得到:
error C2039: 't1' : is not a member of 'Interface<T1,0,_C>
error C2039: 't1' : is not a member of 'Interface<T1,1,_C>
error C2039: 't2' : is not a member of 'Interface<T2,0,_C>
error C2039: 't2' : is not a member of 'Interface<T2,1,_C>
**我已经拆了我的测试代码到解释块,把它们放在一起进行整体测试情况**
这是的“通用”基本模板:
enum eType { T1, T2 };
enum eT1 { T1_I1, T1_I2 };
enum eT2 { T2_I1, T2_I2 };
//This defines the 'global/default' interface that is required
template< eType _T, int _I, typename _C>
struct Interface
{
bool initialise();
};
为此,我部分专门基于所述_T参数模板中添加由INITIALISE使用成员变量()等等:
//t1 has a new member function which initialise() uses
template< int _I, typename _C>
struct Interface< T1, _I, _C >
{
bool initialise();
void t1();
};
//t2 has a new member function which initialise() might uses
template< int _I, typename _C>
struct Interface< T2, _I, _C >
{
bool initialise();
void t2();
};
//We can implement a function for T1 type
template< int _I, typename _C>
bool Interface< T1, _I, _C >::initialise()
{ printf("T1 initialise\n"); return true; }
//We can implement a function for T2 type
template< int _I, typename _C>
bool Interface< T2, _I, _C >::initialise()
{ printf("T2 initialise\n"); return true; }
//We can implement a function for T1 special function
template< int _I, typename _C>
void Interface< T1, _I, _C >::t1()
{ printf("T1\n"); }
//We can implement a function for T2 special function
template< int _I, typename _C>
void Interface< T2, _I, _C >::t2()
{ printf("T2\n"); }
现在到了我无法弄清楚如何根据第二个模板参数_I来专门实现t1()和t2()函数的地方。
//################ ISUE BELOW ###################
//ERROR: We can't implement the special function for T1 based on _I specialization
template< typename _C>
void Interface< T1, (int)T1_I1, _C >::t1()
{ printf("T1_I1 Special function\n"); }
//ERROR: We can't implement the special function for T1 based on _I specialization
template< typename _C>
void Interface< T1, (int)T1_I2, _C >::t1()
{ printf("T1_I2 Special function\n"); }
//ERROR: We can't implement the special function for T2 based on _I specialization
template< typename _C>
void Interface< T2, (int)T2_I1, _C >::t2()
{ printf("T2_I1 Special function\n"); }
//ERROR: We can't implement the special function for T2 based on _I specialization
template< typename _C>
void Interface< T2, (int)T2_I2, _C >::t2()
{ printf("T2_I2 Special function\n"); }
//################ ISUE END ###################
我们在main()函数测试中,它的所有编译:
int _tmain(int argc, _TCHAR* argv[])
{
struct Config {};
Interface<T1, T1_I1, Config> t1i1;
Interface<T1, T1_I2, Config> t1i2;
Interface<T2, T2_I1, Config> t2i1;
Interface<T2, T2_I2, Config> t2i2;
t1i1.initialise();
t1i2.initialise();
t1i1.t1();
t1i2.t1();
t2i1.initialise();
t2i2.initialise();
t2i1.t2();
t2i2.t2();
return 0;
}
的问题会被编译器造成没有看到原始类专业化的存在和它使用的是非专用接口,它没有t1()或t2()。我在哪里得到的语法错误,或者是否有简单的黑客/解决方法来完成我正在尝试做的事情。只要解决方案能够以Serial<UART,Hardware,Config> io
的形式产生一个类型,它就符合我的目标!
你不能做到这一点。函数模板不能被部分地专门化。 – 2013-02-20 23:41:30