[英]The overloaded operator<< template doesn't work for std::list, although it does for std::vector
通過模板,我重載了operator <<,以便它輸出容器的所有元素:
template<typename T, template<typename> typename C>
ostream& operator<<(ostream& o, const C<T>& con) { for (const T& e : con) o << e; return o; }
它可以與std::vector
一起正常工作,但是當我嘗試將其應用於std::list
時會產生錯誤消息:
錯誤:與'operator <<'不匹配(操作數類型為'std :: ostream {aka std :: basic_ostream}'和'std :: __ cxx11 :: list')cout << li;
這是我的代碼摘錄(在GCC 5.2.1,Ubuntu 15.10上編譯):
#include "../Qualquer/std_lib_facilities.h"
struct Item {
string name;
int id;
double value;
Item(){};
Item(string n, int i, double v):
name{n}, id{i}, value{v} {}
};
istream& operator>>(istream& is, Item& i) { return is >> i.name >> i.id >> i.value; }
ostream& operator<<(ostream& o, const Item& it) { return o << it.name << '\t' << it.id << '\t' << it.value << '\n'; }
template<typename T, template<typename> typename C>
ostream& operator<<(ostream& o, const C<T>& con) { for (const T& e : con) o << e; return o; }
int main()
{
ifstream inp {"../Qualquer/items.txt"};
if(!inp) error("No connection to the items.txt file\n");
istream_iterator<Item> ii {inp};
istream_iterator<Item> end;
vector<Item>vi {ii, end};
//cout << vi;//this works OK
list<Item>li {ii, end};
cout << li;//this causes the error
}
但是,當我專門為std::list
編寫模板時,它可以正常工作:
template<typename T>
ostream& operator<<(ostream& o, const list<T>& con) { for (auto first = con.begin(); first != con.end(); ++first) o << *first; return o; }
為什么ostream& operator<<(ostream& o, const C<T>& con)
模板竟然不適用於std::list
?
template<typename T, template<typename> typename C>
ostream& operator<<(ostream& o, const C<T>& con) { for (const T& e : con) o << e; return o; }
為什么這么復雜? 您只需要在您的for循環中使用名稱T
即可。 您也可以通過C::value_type
獲得它,或者只使用auto
關鍵字:
template<typename C>
ostream& operator<<(ostream& o, const C& con)
{
for (const typename C::value_type& e : con) o << e; return o;
}
template<typename C>
ostream& operator<<(ostream& o, const C& con)
{
for (auto& e : con) o << e; return o;
}
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