Variadic模板，没有匹配的函数调用

好吧，这很有效：

#include <iostream>
#include <deque>
#include <tuple>
#include <string>
#include <type_traits>
#include <algorithm>
#include <limits>
template <class ContainerA>
unsigned commonLength(unsigned len, const ContainerA &first) {
  unsigned firstLen = first.size();
  if (len > firstLen) {
    len = firstLen;
  }
  return len;
}

template <class ContainerA, class... Containers>
unsigned commonLength(unsigned len, const ContainerA &first, const Containers&... rest) {
  unsigned firstLen = first.size();
  if (len > firstLen) {
    len = firstLen;
  }
  return commonLength(len, rest...);
}
template <template <typename...> class Container, typename TypeA>
std::tuple<TypeA> getTupleFrom(unsigned index, Container<TypeA> const& first) {
  return std::tuple<TypeA>(first[index]);
}
template <template <typename...> class Container, typename TypeA, typename... Types>
std::tuple<TypeA, Types...> getTupleFrom(unsigned index, Container<TypeA> const& first, Container<Types> const&... rest) {
  return std::tuple_cat(std::tuple<TypeA>(first[index]), getTupleFrom(index, rest...));
}
template <template <typename...> class Container, typename... Types>
Container<std::tuple<Types...>> zip(Container<Types> const&... args) {
  unsigned len = commonLength(std::numeric_limits<unsigned>::max(), args...);
  Container<std::tuple<Types...>> res;
  std::tuple<Types...> item;
  for (unsigned i=0; i<len; i++) {
    item = getTupleFrom(i, args...);
    res.push_back(item);
  }
  return res;
}
int main() {
  std::deque<int> test1 = {1, 2, 3, 4};
  std::deque<std::string> test2 = {"hihi", "jump", "queue"};
  std::deque<float> test3 = {0.2, 8.3, 7, 123, 2.3};
  for (auto i : zip(test1, test2, test3)) {
    std::cout << std::get<0>(i) << std::get<1>(i) << std::get<2>(i) << std::endl;
  }
  //expected output:
  //1hihi0.2
  //2jump8.3
  //3queue7
}

它输出的正是您所期望的。问题是：

• 您没有在commonLength中使用const&容器，而zip的参数在const引用的地方
• commonLength中的无符号参数无法推导，所以我把它移到了开头
• 您以错误的顺序声明/定义了函数（A需要B，但A是在B之前定义的），所以我对它们进行了重新排序

显然，clang 3.1未能推导出zip中的模板参数，但g++4.6可以很好地推导出它们。

逐步拾取。

您缺少标题：

#include <limits>

您有未声明的标识符：

template <template <typename...> class Container, typename TypeA, typename... Types>
std::tuple<TypeA, Types...> getTupleFrom(unsigned index, Container<TypeA> const& first, Container<Types> const&... rest) {
  return std::tuple_cat(std::tuple<TypeA>(first[index]), getTupleFrom(index, rest...), end);
}

end在哪里声明？

您不一致：

Container<std::tuple<Types...>> zip(Container<Types> const&... args) {
  unsigned len = commonLength(args...);
  Container<std::tuple<Types...>> res;

是Container<std::tuple<Types...>>还是Container<Types>？或者这正是你的意思？您的代码有点复杂，只需快速查看即可。

那么，对于Container<TypeA>的非零计数，您只有getTupleFrom的版本，

template <template <typename...> class Container, typename TypeA, typename... Types>
std::tuple<TypeA, Types...> getTupleFrom(unsigned index, Container<TypeA> const& first, Container<Types> const&... rest) {
  return std::tuple_cat(std::tuple<TypeA>(first[index]), getTupleFrom(index, rest...), end);
}
template <template <typename...> class Container, typename TypeA>
std::tuple<TypeA> getTupleFrom(unsigned index, Container<TypeA> const& first) {
  return std::tuple<TypeA>(first[index]);
}

这就是出现错误的原因

error: no matching function for call to ‘getTupleFrom(unsigned int&)’

这表明您以某种方式达到了参数列表为空的点（无符号整数参数除外）。我想你需要防止这种情况发生。