[英]Accessing inner variables of parameter pack in C++
有沒有簡單的方法可以訪問整個參數包的內部成員? 假設您有以下代碼
#include <iostream>
class A {
typedef int type;
constexpr static type C = 5;
};
class B {
typedef unsigned type;
constexpr static type C = 6;
};
template <class T1, class ... TList>
std::ostream & printer(std::ostream & out, T1 t1, TList ... plist) {
out << t1 << " ";
return printer(out, plist...);
}
template <class T1>
std::ostream & printer(std::ostream & out, T1 t1) {
return out << t1 << std::endl;
}
template <class ... TList>
std::ostream & proxy(std::ostream & out, TList ... plist) {
return printer(out, std::forward<TList::type ...>(plist::C ...));
}
int main(int argc, const char * argv[])
{
proxy(std::cout, A(), B());
return 0;
}
我希望代理函數解壓縮plist的成員變量並將其傳遞給打印機。 有沒有一種簡單的方法而又無需遍歷參數列表的方法?
清除代碼中的幾個問題后,我可以對其進行編譯:
A
和B
應該是公開其定義的結構。 printer
放在可變參數的printer
之前 代理現在看起來像這樣:
template <class ... TList>
std::ostream & proxy(std::ostream & out, TList ... plist) {
//leave the forward, plist is a variable pack, so operator.
return printer(out, plist.C ...);
//or:
return printer(out, TList::C...);
}
而且由於C
是靜態constexpr成員,因此您只需保留參數pack:
template <class ... TList>
std::ostream & proxy(std::ostream & out) {
return printer(out, TList::C ...);
}
int main(int argc, const char * argv[])
{
proxy<A,B>(std::cout);
return 0;
}
僅供參考,如果C
是A和B的普通成員變量,則對std::forward
的正確調用應如下所示:
template <class ... TList>
std::ostream & proxy(std::ostream & out, TList&& ... plist) {
return printer(out, std::forward<typename TList::type>(plist.C)...);
}
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