CODEWITHSUNDEEP
c++gccvectorstructtypes
user8685981
user8685981

Reputation:

Type of Vector's Elements for Use in Iterator Type

If I have a struct s:

template<typename T>
struct s {
    std::vector<T> list;
};

Defined in main() as:

s<long double> x;

And I try to get the type of list's iterator using:

typedef typename std::vector<decltype(x.list[0])>::iterator IT;

GCC 7.2.1 produces a bunch of errors, beginning with:

/usr/include/c++/7/bits/alloc_traits.h:392:27: error: forming pointer to reference type 'long double&'
    using pointer = _Tp*;
                        ^

I doubt there is anything wrong with the underlying implementation of C++ in GCC, which this error seems to indicate.

Using this method also assumes that list is not empty, so I instead tried using ::value_type (as suggested on other SO answers):

typedef typename std::vector<x.list::value_type>::iterator IT;

GCC again produces errors:

error: the value of 'x' is not usable in a constant expression.
    typedef typename std::vector<x.list::value_type>::iterator IT;
                                   ^~~~

In order to find an alternative, I added a type-holding variable to the struct:

template<typename T>
struct s {
    T typeholder;
    std::vector<T> list;
};

And this worked fine, using:

typedef typename std::vector<decltype(x.typeholder)>::iterator IT;

Where am I going wrong here? It seems convoluted to need to define an unused variable just to hold the type. Am I misusing ::value_type?

Upvotes: 0

Views: 71

Answers (1)

Mikhail Maltsev
Mikhail Maltsev

Reputation: 1678

The type of decltype(x.list[0]) is long double&. std::vector cannot store references, hence you get an error. The second error ("error: the value of 'x' is not usable in a constant expression.") also seems pretty obvious: you need a type instead of a value. This should compile without errors:

#include <vector>

template<typename T>
struct s {
    std::vector<T> list;
};

int main()
{
    s<long double> x;
    typedef typename std::vector<decltype(x.list)::value_type>::iterator IT;
}

Upvotes: 1

Related Questions