This is my c++ template class:
template <class Type, class Key = ColumnKey>
class Column {
protected:
std::shared_ptr<Type> doGet() const {
std::lock_guard<std::mutex> lock(mutex_);
return std::make_shared<Type>(value_);
}
void doSet(const std::shared_ptr<Type> &value) {
std::lock_guard<std::mutex> lock(mutex_);
value_ = *value;
}
private:
Type value_;
std::mutex mutex_;
};
template<class... Columns>
class Table : private Columns... {
public:
template<class Type, class Key = ColumnKey>
std::shared_ptr<Type> get() {
return Column<Type, Key>::doGet();
}
template<class Type, class Key = ColumnKey>
void set(const std::shared_ptr<Type> &value) {
Column<Type, Key>::doSet(value);
}
};
I can compile properly if I only call the set
methods:
int main() {
Table3 table3;
table3.set<int>(std::make_shared<int>(1));
table3.set<std::string, Key1>(std::make_shared<std::string>("hello_3a"));
table3.set<std::string, Key2>(std::make_shared<std::string>("hello_3b"));
}
where,
struct Key1;
struct Key2;
using Table3 = Table<Column<int>, Column<std::string, Key1>, Column<std::string, Key2>>;
When I call the get
method:
std::cout<<*table3.get<int>()<<std::endl;
std::cout<<*table3.get<std::string, Key1>()<<std::endl;
I get compilation error saying:
error: no matching constructor for initialization of 'std::lock_guard' std::lock_guard lock(mutex_);
But both doGet
and doSet
I have called using same method. Why am I getting a compilation error only when doGet
runs? Is there a mistake the way I am using this?
I am compiling on MacOS 10.13.1 using:
g++ file.cpp -std=c++11 -o file
Aucun commentaire:
Enregistrer un commentaire