Decltype of optional member

Question

I'm trying to get the type from struct member that is in a std::optional<> that is the return type of a member function.

This is a simplified example:

struct Result
{
    int tag;
    int pos;
};

class Dict
{
public:
    std::optional<Result> search(const char *word)
    {
        return Result{ 1,2 };
    }
};

I'd like to be able to do something like this:

int main()
{
    Dict abc;
    decltype(abc.search(const char*)->pos) position;

    return 0;
}

Answer 1

If you pass an actual parameter to search, it will work (along with making search public):

https://wandbox.org/permlink/0Q3mLW7SmQW4QshE

#include <optional>

struct Result
{
    int tag;
    int pos;
};

class Dict
{
public:
    std::optional<Result> search(const char *word)
    {
        return Result{ 1,2 };
    }
};

int main()
{
    Dict abc;
    decltype(abc.search("")->pos) position;

    return 0;
}

The parameter to search doesn't have to be valid (in terms of what your function expects - because it won't actually call it), it just has to be the correct type.

If you want to deal directly with types, and not instances, as your comment suggests, then @Jarod42 points out you can use the following line for your variable declaration:

decltype(std::declval<Dict>().search(std::declval<const char*>())->pos) position;

https://wandbox.org/permlink/kZlqKUFoIWv1m3M3

Though I probably don't need to point out how unreadable a ~70 character variable type is. I think if it were me I would either just use an int, or I would create a type alias for pos, e.g. using ResultPositionType = int; then use that within your Result struct, and again in main.