我正试图以相反的顺序迭代一个可变模板函数的参数包。我的想法是使用尾部递归;"空";模板函数停止递归:
#include <iostream>
template<>
void f() {}
template<int H, int... T>
void f()
{
f<T...>();
std::cout << H << std::endl;
}
int main()
{
f<1,2,3,4,5>();
return 0;
}
然而,上面的代码没有编译:
p.cc:25:8: error: ‘f’ is not a template function void f() {} ^ p.cc: In instantiation of ‘void f() [with int H = 5; int ...T = {}]’: p.cc:30:12: recursively required from ‘void f() [with int H = 2; int ...T = {3, 4, 5}]’ p.cc:30:12: required from ‘void f() [with int H = 1; int ...T = {2, 3, 4, 5}]’ p.cc:36:18: required from here p.cc:30:12: error: no matching function for call to ‘f()’ f<T...>(); ^ p.cc:28:6: note: candidate: template<int H, int ...T> void f() void f() ^ p.cc:28:6: note: template argument deduction/substitution failed: p.cc:30:12: note: couldn't deduce template parameter ‘H’ f<T...>();
我觉得这只是一个语法错误,但我自己无法找到解决方案。知道吗?
因为您应该在专业化之前提供模板声明:
#include <iostream>
template<typename...> void f();
template<>
void f() {}
template<int H, int... T>
void f()
{
f<T...>();
std::cout << H << std::endl;
}
int main()
{
f<1,2,3,42,5>();
return 0;
}
我们开始:https://ideone.com/TZal7p