提取C ++模板参数 [英] Extract C++ template parameters

问题描述

尽管我对此表示怀疑，但我对是否有可能使用RTTI从现有类型中提取原始类型模板参数感到好奇。

Although I'm doubtful, I'm curious as to whether it's possible to extract primitive-type template parameters from an existing type, perhaps using RTTI.

对于例如：

typedef std::bitset<16> WordSet;

是否可以在上面的代码中提取数字16而无需在其他地方进行硬编码？欢迎编译器特定的实现，尽管我对 g ++ 特别感兴趣。

Would it be possible to extract the number 16 in the above code without hard-coding it elsewhere? Compiler specific implementations are welcome, though I'm particularly interested in g++.

推荐答案

不可能。您通常的操作方式是：

It's not possible. The usual way you do it is this:

template<int N>
struct foo {
    static const int value = N;
};

和类型

template<typename T>
struct foo {
    typedef T type;
};

您可以使用 foo< 39> :: value 或 foo< int> :: type 。

如果您有特定类型，则可以使用部分模板特化：

If you have a particular type, you can use partial template specialization:

template<typename>
struct steal_it;

template<std::size_t N>
struct steal_it< std::bitset<N> > {
    static const std::size_t value = N;
};

类型参数的确可以使用相同的原理。现在您可以将任何位传递给它，例如 steal_it< std :: bitset< 16> > :: value （请注意使用size_t，而不是int！）。由于我们尚无可变参数的许多模板参数，因此我们必须将自己限制为特定的参数计数，并对数量从1到N的数量重复重复secret_it模板专门化。另一个困难是扫描具有混合参数的类型（类型和非参数类型）。类型参数）。这可能是不容易解决的。

The same principle is possible for type parameters too, indeed. Now you can pass any bitset to it, like steal_it< std::bitset<16> >::value (note to use size_t, not int!). Because we have no variadic many template paramters yet, we have to limit ourself to a particular parameter count, and repeat the steal_it template specializations for count from 1 up to N. Another difficulty is to scan types that have mixed parameters (types and non-types parameters). This is probably nontrivial to solve.

如果没有类型，而只是对象，则可以使用技巧，在编译时仍获得值：

If you have not the type, but only an object of it, you can use a trick, to still get a value at compile time:

template<typename T>
char (& getN(T const &) )[steal_it<T>::value];  

int main() {
    std::bitset<16> b;
    sizeof getN(b); // assuming you don't know the type, you can use the object
}

技巧是使函数模板自动推导类型，然后返回对字符数组的引用。该函数不需要定义，唯一需要的就是它的类型。

The trick is to make the function template auto-deduce the type, and then return a reference to a character array. The function doesn't need to be defined, the only thing needed is its type.

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