可选模板参数组合
我有一组模板专门化(C++11)，它使用下面的基本模板，其中同时存在tPayLoad和tReturnType：
template <class tDerivedOperation, class tBridgeType, class tPayLoadType = void, class tReturnType = void> class OperationT ...
第一个专门化是在没有tPayLoadType或tReturnType的情况下。
template <class tDerivedOperation, class tBridgeType> class OperationT<tDerivedOperation, tBridgeType, void, void> ...
第二个专门化是当只有tPayLoadType时。
template <class tDerivedOperation, class tBridgeType, class tPayLoadType> class OperationT<tDerivedOperation, tBridgeType, tPayLoadType, void> ...
最后，我想要第三种情况(第四种组合)，没有tPayLoadType，只有一个tReturnType，这就是问题出现的地方。我不能用(我想)：
template <class tDerivedOperation, class tBridgeType, class tReturnType> class OperationT<tDerivedOperation, tBridgeType, void, tReturnType> ...
因为这在本质上和专门化是一样的。如果可能的话，有谁知道怎么做呢？
/N
回答 1
Stack Overflow用户
发布于 2021-11-09 21:03:48
以下是
#include <iostream>
template<class tDerivedOperation, class tBridgeType, class tPayLoadType = void, class tReturnType = void>
struct OperationT {
static constexpr auto overload = "base";
};
template<class tDerivedOperation, class tBridgeType>
struct OperationT<tDerivedOperation, tBridgeType, void, void> {
static constexpr auto overload = "both void";
};
template<class tDerivedOperation, class tBridgeType, class tPayLoadType>
struct OperationT<tDerivedOperation, tBridgeType, tPayLoadType, void> {
static constexpr auto overload = "back void";
};
template<class tDerivedOperation, class tBridgeType, class tReturnType>
struct OperationT<tDerivedOperation, tBridgeType, void, tReturnType> {
static constexpr auto overload = "front void";
};
int main() {
std::cout << OperationT<int, int>::overload << '\n';
std::cout << OperationT<int, int, int>::overload << '\n';
std::cout << OperationT<int, int, void, int>::overload << '\n';
std::cout << OperationT<int, int, int, int>::overload << '\n';
}给出输出
both void
back void
front void
base就像你所期望的那样。这能回答你的问题吗？
https://stackoverflow.com/questions/69904573
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