c++ - 递归、可变参数模板函数

Question

我知道我可以使用常规的可变参数函数来做到这一点，但我想用模板来做到这一点。我的 (C++17) 编译器不同意。

#include <cstdint>

unsigned int fct(unsigned int k)
{
    return k;
}

template<std::size_t First, std::size_t... Other>
unsigned int fct(unsigned int k)
{
    return First * fct<Other...>(k);
}

int main(void)
{
    //const auto k = fct<3u, 5u, 7u>(2u);
    return 0;
}

上面的代码编译得很好。但是，如果我选择取消注释的声明k，编译将失败并显示以下报告：

foo.cpp: In function ‘int main()’:
foo.cpp:17:13: warning: unused variable ‘k’ [-Wunused-variable]
  const auto k = fct<3u, 5u, 7u>(2u);
             ^
foo.cpp: In instantiation of ‘unsigned int fct(unsigned int) [with     long unsigned int First = 7; long unsigned int ...Other = {}]’:
foo.cpp:11:30:   recursively required from ‘unsigned int fct(unsigned int) [with long unsigned int First = 5; long unsigned int ...Other = {7}]’
foo.cpp:11:30:   required from ‘unsigned int fct(unsigned int) [with long unsigned int First = 3; long unsigned int ...Other = {5, 7}]’
foo.cpp:17:35:   required from here
foo.cpp:11:30: error: no matching function for call to ‘fct<>(unsigned int&)’
  return First * fct<Other...>(k);
                 ~~~~~~~~~~~~~^~~
foo.cpp:9:14: note: candidate: template<long unsigned int First, long unsigned int ...Other> unsigned int fct(unsigned int)
 unsigned int fct(unsigned int k)
              ^~~
foo.cpp:9:14: note:   template argument deduction/substitution failed:
foo.cpp:11:30: note:   couldn't deduce template parameter ‘First’
  return First * fct<Other...>(k);
                 ~~~~~~~~~~~~~^~~

我怎样才能使这项工作？我认为创建一个非模板fct()就可以了。

标签： c++templatesvariadic-templatesc++17template-meta-programming

Answer 1

由于问题标记为 C++17，您可能想尝试新引入的折叠表达式( live )。

#include <cstdint>
#include <iostream>

template <std::size_t... Args>
auto fct(std::size_t k) noexcept {
  return (k * ... * Args);
}

int main() {
  std::cout << fct<>(1) << std::endl;
  std::cout << fct<2, 3>(1) << std::endl;
}