我想知道为什么在我的示例中我不能使用set supportList to choose from list {"A", "B"} with prompt "Source:" default items {"A"}
if supportList is false then return
if supportList is {"A"} then
set responseList to {"1", "2", "3", "4"}
else
set responseList to {"5", "6", "7", "8", "9", "10"}
end if
set misrouteList to choose from list responseList with title "Prepared Responses" with prompt "Reason:"
if misrouteList is false then return
set supportChoice to item 1 of supportList
set prChoice to item 1 of misrouteList
说明符来选择特殊的模板方法。
仅当我使用声明的模板参数来指向专业化时，代码才能按预期工作：
decltype不出所料，字符串参数的专业化被称为一次字符串参数。
但是当我使用template <typename T>
auto sum(const T& value)
{
std::cout << "sum for template" << std::endl;
return sizeof(value);
}
template<>
auto sum(std::string const& value)
{
std::cout << "sum for string" << std::endl;
return value.length();
}
template <typename Last>
auto sumBytes(const Last& last)
{
return sum<Last>(last);
}
template <typename First, typename ...Tail>
auto sumBytes(const First& first, const Tail& ...tail)
{
return sum<First>(first) + sumBytes(tail...);
}
int main()
{
std::string str = "hello";
auto sum = sumBytes(str,2,3,4);
}
确定第一个参数的类型时，没有调用字符串的专用函数，而是选择了通用函数：
decltype我想知道为什么在使用template <typename T>
auto sum(const T& value)
{
std::cout << "sum for template" << std::endl;
return sizeof(value);
}
template<>
auto sum(std::string const& value)
{
std::cout << "sum for string" << std::endl;
return value.length();
}
template <typename Last>
auto sumBytes(const Last& last)
{
return sum<decltype(last)>(last);
}
template <typename First, typename ...Tail>
auto sumBytes(const First& first, const Tail& ...tail)
{
return sum<decltype(first)>(first) + sumBytes(tail...);
}
int main()
{
std::string str = "hello";
auto sum = sumBytes(str,2,3,4);
}
说明符时，不调用字符串的专门化方法？据我所知，decltype应该返回decltype类型。
答案 0 :(得分：2)
decltype(first)将产生std::string const &，而函数专门用于T为std::string的情况。可以通过添加简历限定符和参考来解决此问题：
return sum<::std::remove_const_t<::std::remove_reference_t<decltype(first)>>>(first) + sumBytes(tail...);