sag*_*aga 1 c++ templates metaprogramming template-meta-programming
我正在尝试通过实现一些功能来学习 C++ 模板元编程。我知道已经在 stackoverflow 上提供了这个特定问题的解决方案,我感兴趣的是了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;
}
Run Code Online (Sandbox Code Playgroud)
这个模板应该检查一个元组是否包含某种类型的元素,但是编译它会出现以下错误:
> 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.
Run Code Online (Sandbox Code Playgroud)
这段代码有什么问题?
由于您在代码中使用 c++17,我认为有必要指出有很多新工具可以避免必须制作此类递归模板。
你可以把整个事情浓缩成这样:
#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;
}
Run Code Online (Sandbox Code Playgroud)
如果std::disjunction没有参数,它默认为 false,所以这里也介绍了传递一个空元组。