c++ - variadic template for sorting variable number of arrays

Question

I am trying to write a program to sort unsorted 'n' vectors using variadic templates into one sorted vector. It can be done in other ways but I want to do it this way to better my understanding of variadic templates. I don't know if I can achieve that but below is my attempt at it which is not working.

My understanding is that ultimately sorted_n_vector will be reduced to MergeVector<int> + MergeVector<int> + MergeVectorMergeVector<int> same as line 50(MergeVector a8(a5 + a6 + a7)) but I think it is wrong. Please help and also give me some code review comments as well.

#include <iostream>
#include <vector>
#include <iterator>
#include <algorithm>
#include <functional>
#include <string>

template<typename t>
class MergeVector {
        public:
        std::vector<t> arr_;

        MergeVector(std::vector<t> arr) : arr_(arr) {}

        auto operator + (MergeVector<t> &obj) {
                std::vector<t> dst;
                std::sort(arr_.begin(), arr_.end());
                std::sort(obj.arr_.begin(), obj.arr_.end());
                std::merge(arr_.begin(), arr_.end(), obj.arr_.begin(), obj.arr_.end(), std::back_inserter(dst));
                MergeVector res(dst);
                return res;
        }

        friend auto& operator<<(std::ostream &os, MergeVector<t>& mv)
        {
                std::copy(mv.arr_.begin(), mv.arr_.end(), std::ostream_iterator<t>(os, " "));
                return os;
        }
};

template<typename t>
auto sorted_n_vector(t vec)
{
        return vec;
}

template<typename t, typename... vectors>
auto sorted_n_vector(t vec, vectors... args)
{
        return vec + sorted_n_vector(args...);
}

int main() {
        std::vector<int> a1 = {1, 2, 3};
        std::vector<int> a2 = {4, 5, 6};
        std::vector<int> a3 = {7, 8, 9};
        MergeVector<int> a5(a1);
        MergeVector<int> a6(a2);
        MergeVector<int> a7(a3);
        MergeVector<int> a8(a5 + a6 + a7); //---- Line 50
        std::cout << a8 << std::endl;

#if 1
        MergeVector<int> a9(sorted_n_vector(a5, a6, a7));
        std::cout << a9 << std::endl;
#endif

        return 0;
}

I am getting below error:

test.cpp:40:20: error: invalid operands to binary expression ('MergeVector<int>' and 'MergeVector<int>')
        return vec + sorted_n_vector(args...);
               ~~~ ^ ~~~~~~~~~~~~~~~~~~~~~~~~
test.cpp:40:22: note: in instantiation of function template specialization 'sorted_n_vector<MergeVector<int>, MergeVector<int> >' requested here
        return vec + sorted_n_vector(args...);
                     ^
test.cpp:54:22: note: in instantiation of function template specialization 'sorted_n_vector<MergeVector<int>, MergeVector<int>, MergeVector<int> >' requested here
        MergeVector<int> a9(sorted_n_vector(a5, a6, a7));
                            ^
test.cpp:15:7: note: candidate function not viable: expects an l-value for 1st argument
        auto operator + (MergeVector<t> &obj) { 

标签： c++templatesvariadic-templates

Answer 1

sorted_n_vector具有指定为的返回类型auto，因此它将返回一个临时对象，而不是对其他对象的引用。

您尝试operator+将此临时作为参数调用，但operator+需要一个非常量（左值）引用参数。

禁止将临时变量（或者更确切地说是右值表达式）绑定到非 const（左值）引用，因此编译器不会考虑调用您operator+，而是会在其他地方搜索。

没有找到任何替代调用+，它会为您提供您所看到的错误消息。

解决方法是将参数operator+作为 const 引用：

auto operator + (const MergeVector<t> &obj) {

将临时变量（右值表达式）绑定到 const（左值）引用是允许的，并且可以按预期工作。

编辑：

但是，这会引发另一个错误，因为您operator+实际上是在修改objand this。虽然您可以通过按值而不是按引用来解决这个问题，但这是一个不寻常的设计。通常+预计不会修改其参数。如果c = a+b因为a或b不允许修改而失败，那将是令人惊讶的。您可以通过在MergeVector构造函数中进行排序来实现这一点：

MergeVector(std::vector<t> arr) : arr_(arr) {
    std::sort(arr_.begin(), arr_.end());
}

然后你可以把这些std::sort线放进去，你可以在两个参数中operator+完全做到：const

auto operator + (const MergeVector<t> &obj) const {