检查元组是否包含某种类型的元素

时间:2018-03-19 04:15:17

标签: c++ templates metaprogramming template-meta-programming

我试图通过实现一些功能来学习c ++模板元编程。我知道在stackoverflow上提供了这个特定问题的解决方案,我感兴趣的是理解为什么这个解决方案不起作用。这是代码:

template < std::size_t... Ns , typename... Ts >
auto tail_impl( std::index_sequence<Ns...> , std::tuple<Ts...> t )
{
   return  std::make_tuple( std::get<Ns+1u>(t)... );
}

template <class F, class... R >
tuple<R...> tail( std::tuple<F,R...> t )
{
   return  tail_impl( std::make_index_sequence<sizeof...(R)>() , t );
}


template<class X, class F, class... R>
constexpr bool check_for_type(tuple<F,R...> t) {
    if constexpr(is_same<F,X>::value) {
        return true;
    } 
    return check_for_type<X>(tail(t));
}


template<class X>
constexpr bool check_for_type(tuple<> t) {
    return false;
}


int main( int argc, const char *argv) {
    auto t2 = make_tuple(4,"qw", 6.5);
    double f = check_for_type<double>(t2);
    return 0;
}

该模板应该检查元组是否包含某种类型的元素,但是编译它会产生以下错误:

> clang++ t.cpp -std=c++17
t.cpp:45:12: error: call to function 'check_for_type' that is neither visible in the
      template definition nor found by argument-dependent lookup
    return check_for_type<X>(tail(t));
           ^
t.cpp:45:12: note: in instantiation of function template specialization
      'check_for_type<double, double>' requested here
t.cpp:45:12: note: in instantiation of function template specialization
      'check_for_type<double, const char *, double>' requested here
t.cpp:66:16: note: in instantiation of function template specialization
      'check_for_type<double, int, const char *, double>' requested here
    double f = check_for_type<double>(t2);
               ^
t.cpp:58:16: note: 'check_for_type' should be declared prior to the call site
constexpr bool check_for_type(tuple<> t) {
               ^
1 error generated.

这段代码有什么问题?

2 个答案:

答案 0 :(得分:4)

由于您在代码中使用了c ++ 17,我认为有必要指出有很多新工具可以避免制作这些递归模板。

你可以将整个事情浓缩成:

HandleScope

如果#include <iostream> #include <type_traits> #include <tuple> template <typename T1, typename... T2> constexpr bool check_for_type(std::tuple<T2...>) { return std::disjunction_v<std::is_same<T1, T2>...>; } int main() { std::tuple<int, char, bool> tup; std::cout << check_for_type<char>(tup) << '\n'; std::cout << check_for_type<float>(tup) << std::endl; return 0; } 没有参数,则默认为false,因此这里也会介绍传递空元组。

答案 1 :(得分:2)

template<class X, class F, class... R>
constexpr bool check_for_type(tuple<F,R...> t) {
    if constexpr(is_same<F,X>::value) {
        return true;
    } 
    return check_for_type<X>(tail(t));
}

您在宣布之前呼叫check_for_type<X>(...)。有帮助的是,您的错误消息指出:

t.cpp:58:16: note: 'check_for_type' should be declared prior to the call site
constexpr bool check_for_type(tuple<> t) {

执行此操作后,代码compiles

// Put this function first
template<class X>
constexpr bool check_for_type(tuple<> t) {
    return false;
}

template<class X, class F, class... R>
constexpr bool check_for_type(tuple<F,R...> t) {
    if constexpr(is_same<F,X>::value) {
        return true;
    }
    // Right here, the compiler looks up `check_for_type` and doesn't find
    // an overload that can be called with just one explicit type parameter.
    return check_for_type<X>(tail(t));
}