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I was trying to loop a list and getting the address of the element. But all of the element's address is the same when using variant class.

int main()
{
    std::list<std::variant<std::list<int>, std::list<std::string>>> list_of_lists;
    std::list<int> int_list = {1, 2, 3};
    string a1 ="one";
    std::list<std::string> string_list = {a1, "two", "three"};

    list_of_lists.push_back(int_list);
    list_of_lists.push_back(string_list);

    // Iterate over all std::variant elements.
    for (const auto& v : list_of_lists) {
        // Check if the variant holds a list of ints.
        if (std::holds_alternative<std::list<int>>(v)) {
            // 'v' holds a list of ints.
            for (auto i : std::get<std::list<int>>(v)) {
                std::cout << i << ' ';
                cout<<&i<<endl;
            }
            std::cout << '\n';
        }
            // Check if the variant holds a list of strings.
        else if (std::holds_alternative<std::list<std::string>>(v)) {
            // 'v' holds a list of strings.
            for (const auto& s : std::get<std::list<std::string>>(v)) {
                std::cout << s << ' ';
                cout<<"string: "<<&std::get<std::list<std::string>>(v)<<endl;
                cout<<&s<<endl;
            }
            std::cout << '\n';
        }
    }
    cout<<"a1 address: "<<&a1<<endl;
}
output:
1 0x8dfcac
2 0x8dfcac
3 0x8dfcac

one string: 0x8f2890
0x8f27d0
two string: 0x8f2890
0x8f2810
three string: 0x8f2890
0x8f2850

a1 address: 0x8dfcc0

As you ## Heading ##

It looks like the variant std library cannot get the original variable's own address. Thus, every time I try to loop through the linked list, all the address of elements in the mother list are changed. Thus, I can only display and I can not push back for those linked lists in the mother linked list. One of my friends told me c++ is all about pointer and he hates Java because it does have pointer! And now I can not use a pointer in c++ with this library. I have heard this problem can be solved by boot:: variant, std:: optional or polymorphism. But I have no idea how these work. Anyway, some people said the reason this library does not touch the address idea is because they have a disagreement about how what assignment

Jonathan Sum
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1 Answers1

1

For the int loop, you created a local variable and take its address.

for (auto i : std::get<std::list<int>>(v))  // local variable, same address each time

should be

for (auto& i : std::get<std::list<int>>(v)) // reference, refers to the actual element

This has nothing to do with variants.

For the string loop, you already use a reference. 

for (const auto& s : std::get<std::list<std::string>>(v)) // reference

That's why things work as expected.

L. F.
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  • Nope, It is still wrong. Even though I used & s, I am still unable to get the original address. – Jonathan Sum Jun 02 '19 at 03:03
  • I can not see anything in the link you posted. It just keeps loading forever. – Jonathan Sum Jun 02 '19 at 03:17
  • @ChannelNews I can. ([screenshot](https://i.stack.imgur.com/5Wysf.png)) Maybe there is something wrong with your internet connection, but I am not sure. – L. F. Jun 02 '19 at 03:20
  • I just checked. No, It does not work. Once they are in the loop, all elements' address are not the same. They are just different. – Jonathan Sum Jun 02 '19 at 03:25
  • @ChannelNews If I understood it correctly, your question asks about why the addresses are the same. – L. F. Jun 02 '19 at 03:26
  • My title is "unable to get the address of the element in the list by variant". Thus, my question is about the address of the element in the list by variant. – Jonathan Sum Jun 02 '19 at 03:27
  • @ChannelNews A question begins with a word like "what" or "why" or is a yes/no question. Your question is a statement. A statement is not a question. – L. F. Jun 02 '19 at 03:30
  • I think anyone can understand the title by what it means. – Jonathan Sum Jun 02 '19 at 04:54