在下面的代码中,initialize()说明了一个基于编译时多态的方法.编译的initialize()版本取决于int2type< true>和int2type< false>,对于给定的模板参数T,其中只有一个为真.
它恰好发生在数据成员T * m_datum;将适用于int2type< true>和int2type< false>.
现在,我想更改int2type< false>版本到std :: vector< T> m_datum;,所以我的问题是,我如何修改我的代码,以便数据成员m_datum在int2type<>？上是多态的？
注意：请忽略下面代码背后的基本原理 – 相反,我想重点关注为数据成员实现编译时多态的机制.
#include <type_traits>
#include <stdlib.h>
using namespace std;
template <bool n>
struct int2type
{
enum { value = n };
};
template< typename T >
struct is_trivially_copyable
{
static const bool value = std::is_standard_layout<T>::value;
};
template<class T>
class Foo
{
public:
Foo( size_t n ) : m_nr( n )
{
initialize( int2type<is_trivially_copyable<T>::value>() );
}
~Foo() { }
private:
void initialize( int2type<true> )
{
m_datum = (T*) calloc( sizeof(T),m_nr );
}
void initialize( int2type<false> )
{
m_datum = new T[m_nr];
}
private:
size_t m_nr;
T* m_datum; // ok for int2type<true>
// vector<T> m_datum; // want to change to this for int2type<false>
};
class Bar
{
public:
Bar() { }
virtual ~Bar() { }
};
int main(int argc,char** argv)
{
Foo<int> foo_trivial( 5 );
Foo<Bar> foo_nontrivial( 10 );
return 0;
}
C 11解决方案,基于Nawaz的建议
#include <type_traits>
#include <vector>
#include <stdlib.h>
using namespace std;
template< typename T >
struct is_trivially_copyable
{
static const bool value = std::is_standard_layout<T>::value;
};
template<class T>
class Foo
{
private:
static const bool what = is_trivially_copyable<T>::value;
typedef typename std::conditional<what,T*,std::vector<T>>::type type;
public:
Foo( size_t n ) : m_nr( n )
{
initialize( m_datum );
}
~Foo() { }
private:
void initialize( T* dummy )
{
m_datum = (T*) calloc( sizeof(T),m_nr );
}
void initialize( std::vector<T>& dummy )
{
m_datum.resize( m_nr );
}
private:
size_t m_nr;
type m_datum;
};
class Bar
{
public:
Bar() { }
virtual ~Bar() { }
};
int main(int argc,char** argv)
{
Foo<int> foo_trivial( 5 );
Foo<Bar> foo_nontrivial( 10 );
return 0;
}
解决方法
C 11解决方案
使用std::conditional作为：
#include <type_traits>
template<class T>
class Foo
{
//some info we can use throughout the class
static const bool what = is_trivially_copyable<T>::value;
typedef typename std::conditional<what,std::vector<T>>::type data_type;
//data members
data_type m_data; //this is what you need!
}
C 03解决方案
您可以编写一个元函数,并将其部分特化如下：
template<class T>
class Foo
{
//primary template
template<bool b,typename T>
struct get { typedef T* type; };
//partial specialization
template<typename T>
struct get<false,T> { typedef std::vector<T> type; };
//some info we can use throughout the class
static const bool what = is_trivially_copyable<T>::value;
typedef typename get<what,T>::type data_type;
//data members
data_type m_data; //this is what you need!
};
因此,当这是真的时,data_type将变成T *,否则它将是std :: vector< T>,如所希望的那样.
