C++中的元组向量

Question

我已经尝试了所有我能做的事情，但是这段代码给了我错误。 两种语法都不起作用。我已经评论了 operator[] 但也请提供解决方案。

#include <bits/stdc++.h>
using namespace std;
int main() {
    typedef vector< tuple<int, int, int> > my_tuple;
    my_tuple tl;
    tl.push_back( tuple<int, int, int>(21,20,19) );
    for (my_tuple::const_iterator i = tl.begin(); i != tl.end(); ++i) {
        //cout << get<0>(tl[i]);
        //cout << get<0>(tl[i]);
        //cout << get<0>(tl[i]);
        cout << get<0>(tl.at(i));
        cout << get<1>(tl.at(i));
        cout << get<2>(tl.at(i));
    }

    return 0;
}

在 for 循环中打印元组时出现错误。

error: no matching function for call to 'std::vector<std::tuple<int, int, int> >::at(std::vector<std::tuple<int, int, int> >::const_iterator&)'

对于运算符[]

error: no match for 'operator[]' (operand types are 'my_tuple {aka std::vector<std::tuple<int, int, int> >}' and 'std::vector<std::tuple<int, int, int> >::const_iterator {aka __gnu_cxx::__normal_iterator<const std::tuple<int, int, int>*, std::vector<std::tuple<int, int, int> > >}')

Answer 1

您的i 是一个迭代器，它有点像一个指针，所以您需要取消引用它，而不是将它传递给operator [] 或at()：

get<0>(*i);

Answer 2

#include <bits/stdc++.h>
using namespace std;
int main() {
    typedef vector< tuple<int, int, int> > my_tuple;
    my_tuple tl; 
    tl.push_back( tuple<int, int, int>(21,20,19) );
    for (my_tuple::const_iterator i = tl.begin(); i != tl.end(); ++i) {
        cout << get<0>(*i) << endl;
        cout << get<1>(*i) << endl;
        cout << get<2>(*i) << endl;
    }
    cout << get<0>(tl[0]) << endl;
    cout << get<1>(tl[0]) << endl;
    cout << get<2>(tl[0]) << endl;

    return 0;
}

Answer 3

#include <bits/stdc++.h>
using namespace std;
int main() {
    vector<tuple<int,string,int>> vt;
    vt.push_back({421,"cha",10});
    vt.push_back({464,"sam",20});
    vt.push_back({294,"sac",30});
    for(const auto &i : vt)
        cout<<get<0>(i)<<"  "<<get<1>(i)<<"  "<<get<2>(i)<<endl;
return 0;
}

Answer 4

为什么不使用std::tie？

#include <vector>
#include <tuple>
#include <iostream>

int main() {
    std::vector<std::tuple<int, int, int>>  tuples;

    tuples.push_back(std::make_tuple(21, 20, 19));
    tuples.push_back(std::make_tuple(18, 17, 19));

    for (auto&& tuple: tuples)
    {
      int X, Y, Z;
      std::tie(X, Y, Z) = tuple;

      std::cout << X << " " << Y << " " << Z << std::endl;
    }

    return 0;
}

或在 c++17 中使用 auto [ var1, var2, ..., varX]

#include <vector>
#include <tuple>
#include <iostream>

int main() {
    std::vector<std::tuple<int, int, int>>  tuples;

    tuples.push_back(std::make_tuple(21, 20, 19));
    tuples.push_back(std::make_tuple(18, 17, 19));

    for (auto [ X, Y, Z ] : tuples)
    {
      std::cout << X << " " << Y << " " << Z << std::endl;
    }

    return 0;
}

Answer 5

vector<tuple<int, int>> my_vec{
    tuple<int, int> { 1, 15 },
    tuple<int, int> { 2, 100 }
};

for(const auto &i : myvec)
    cout<<get<0>(i)<<"  "<<get<1>(i)<<"  "<<get<2>(i)<<endl;