名称和派生名称 class 之间的串联（作为模板参数）

Question

我正在尝试避免此解决方案出现问题：

static const int  FOO_Test = 7;

template < typename Derived >
class Foo {

  public:
    static const int  Type;
};

const int  Foo::Type = FOO_##Derived ;

class Test : public Foo<Test> {};

如您所见，我正在尝试获取 FOO_Test 值，该值仅在有人从 Foo 中派生 class 时才存在（需要一些外部工具来编写 header ).

好吧，宏连接不起作用（毕竟不确定），有实现它的想法吗？

Answer 1

如果你能用C++14，就用skypjack的方法。如果你不能，你可以写这样的东西：

#define DEFINE_FOO_TYPE(Derived) \
    const int Foo<Derived>::Type = FOO_##Derived

DEFINE_FOO_TYPE(Test);

但我会尽可能避免它。

完整示例：

// This probably wants to go in Foo.h
template < typename Derived >
class Foo {

  public:
    static const int  Type;
};
#define DEFINE_FOO_TYPE(Derived) \
    template <> const int Foo<Derived>::Type = FOO_##Derived

// This probably wants to go in Test.h
static const int  FOO_Test = 7;
class Test : public Foo<Test> {};

// This needs to go in Test.cpp (or some other central place)
// Note that it must follow definition of Test.
DEFINE_FOO_TYPE(Test);

// an empty main so the whole example will compile and run.
int main() 
{}

Answer 2

自 C++14 起，您可以使用变量模板来执行此操作。
它遵循一个最小的工作示例：

class Test;

template<typename> static constexpr int FOO;
template<> constexpr int FOO<Test> = 7;

template <typename Derived>
struct Foo {
    static const int Type;
};

template<typename Derived>
const int Foo<Derived>::Type = FOO<Derived> ;

class Test : public Foo<Test> {};

int main () {
    static_assert(Test::Type == 7, "!");
}

它有助于将 FOO 值与 Foo class 值分开。
否则，您可以完全专业化并丢弃这些变量。
例如：

class Test;

template <typename Derived>
struct Foo {
    static const int Type;
};

template<>
const int Foo<Test>::Type = 7 ;

class Test : public Foo<Test> {};

int main () {
    static_assert(Test::Type == 7, "!");
}

名称和派生名称 class 之间的串联（作为模板参数）

Concatenation between name and derived name class (as template args)

c++

templates

concatenation

crtp