这可以与带有模板元编程的字符数组一起使用吗? [英] Is this possible to work with character arrays with Template Metaprogramming?

问题描述

我编写了一系列模板，以基于多达32种数据类型填充32个字符串.是否可以在编译时进行计算?

我的假设是不，但想验证以防万一我错了.

生成具有3种数据类型的注册字符串的代码示例

 temp [0] = Char_Type <   T1  <   T2  > ();
temp [2] = Char_Type <   T3  > ();
return temp; 

解决方案

也许这是一些值得深思的事情. 内联 char char_type(){返回 ' -'; } 模板<> 内联 char char_type< char>(){>返回 ' c'; } 模板<> 内联 char char_type< int>(){返回 ' d'; } 模板< typename T1， typename T2> 类 MyClass { 公共: 静态 字符 temp []; // 将在编译时初始化 }; 模板< typename T1， typename T2> char MyClass< T1，T2> :: temp [] = {char_type< T1>()，char_type< T2>()， int main() { cout<< MyClass< int，char> :: temp<<恩德尔 cin.get(); 返回 0 ; }

将其变成一个有效的示例，并删除了模板参数上的约束.

我不完全确定您要实现的目标，但是由于每个是唯一类型，如果它们是相同基类(或该基类的实例)的子类，则只能将它们分配给相同的集合(在您的情况下为temp[]数组). >

  class  some_base
{
};
模板< T类>  class  Char_Type: public  some_base
{
公共:
    T 值;
    Char_Type()
    {
    };
    Char_Type( const  T& v)
    {
        值 = v;
    };
};
 int  main( int  argc， char  * argv [])
{
    some_base * a = 新 some_base [ 3 ];
    a [ 0 ] = Char_Type< char>('  c ');
    a [ 1 ] = Char_Type< int>( 123 );
    a [ 2 ] = Char_Type< int>( 123 . 23 )；
    删除 [] a;
    返回  0 ;
} 

但是我认为这样的设置会取消您尝试执行的操作.

I''ve written a series of templates to fill in a 32 character string based on up to 32 data types. Is it possible to compute this at compile time?

My assumption is no, but wanted to verify in case I was wrong.

Example of code to generate registration string with 3 data types

temp[0]=Char_Type<T1>();
temp[1]=Char_Type<T2>();
temp[2]=Char_Type<T3>();
return temp;

解决方案

Maybe this would be some food for thought

template<typename T> inline char char_type()       { return '-'; }
template<>           inline char char_type<char>() { return 'c'; }
template<>           inline char char_type<int>()  { return 'd'; }

template <typename T1, typename T2>
class MyClass
{
public:
    static char temp[]; // Will be init'ed at compile time
};

template <typename T1, typename T2>
char MyClass<T1, T2>::temp[] = { char_type<T1>(), char_type<T2>(), 0 };

int main()
{
    cout << MyClass<int, char>::temp << endl;
    cin.get();
    return 0;
}

Edit: Turned it into a working sample, and removed the constraints on the template parameters.

I''m not entirely sure what you''re trying to achieve, but since each Char_Type<T> is a unique type, they can only be assigned to the same collection (in your case the temp[] array), if they''re subclasses of the same base class (or an instance of that base class).

class some_base
{
};
template<class T> class Char_Type : public some_base
{
public:
    T value;
    Char_Type()
    {
    };
    Char_Type(const T& v)
    {
        value = v;
    };
};
int main(int argc, char* argv[])
{
    some_base* a = new some_base[3];
    a[0] = Char_Type<char>('c');
    a[1] = Char_Type<int>(123);
    a[2] = Char_Type<int>(123.23);
    delete[] a;
    return 0;
}

But I think such a setup would cancel what you''re trying to do.

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