Is it possible to construct a fixed sized array in-place as a function argument?

Question

This is mainly a question about a corner of C++ syntax related to fixed-sized arrays.

Suppose I have a function that exploits type information, for example:

template<class T> void fun(T const& t){
    std::cout << typeid(t).name() << std::endl;
}

I can pass a value or a temporary object:

int i;
fun(i); // prints "int" ("i" actually)
fun(int{});   // prints "int" ("i" actually)

However I can't do the same with arrays

double a[10][10];
fun(a); // ok, prints "a[10][10]" ("A10_A10_d" actually)

fun(double[10][10]); // doesn't compile
fun(double{}[10][10]); // doesn't compile
fun(double[10][10]{}); // doesn't compile
fun(double()[10][10]); // doesn't compile
fun(double[10][10]()); // doesn't compile
fun(double(&)[10][10]); // doesn't compile
fun(double(*)[10][10]); // doesn't compile

I could in principle do:

typedef double a1010[10][10];
fun(a1010{});

but, is it possible to do without predefining a typedef?

Is it possible at all to construct a fixed sized array in-place as a function argument?

Full code:

template<class T> void fun(T const& t){
    std::cout << typeid(t).name() << std::endl;
}

typedef double a1010[10][10];

int main(){
    int i;
    fun(i); // prints "int" ("i" actually)
    double a[10][10];
    fun(a); // prints "a[10][10]" ("A10_A10_d" actually)
    fun(a1010{});

    fun(int{});   // prints "int"
/*  fun(double[10][10]); // doesn't compile
    fun(double{}[10][10]); // doesn't compile
    fun(double[10][10]{}); // doesn't compile
    fun(double()[10][10]); // doesn't compile
    fun(double[10][10]()); // doesn't compile
    fun(double(&)[10][10]); // doesn't compile
    fun(double(*)[10][10]); // doesn't compile
    */
    return 0;
}

---

Bonus points (probably a bounty): What about variable-sized arrays?

int N = 10;
f(double[N]);

Answer 1

Try:

fun((int[3]){1,2,3});
fun((int[5]){});

As for the "bonus points": variable sized arrays are not part of the language. This extension to the language does not work with template arguments:

prog.cc:4:6: note: candidate template ignored: substitution failure : variably modified type 'int [n]' cannot be used as a template argument fun(const T&t)

Edit

As Chris noted, the above solution proposes to use compound literals, which are an extension to C++. There is a solution that avoids this extension to C++, using a simple helper class:

template <class T, std::size_t N>
struct my_array
{
    T data[N];
};

template <class T, std::size_t N>
void print(const T (&x)[N])
{
     for (auto i: x)
         std::cout << i << '\n';
}

int main()
{
    print(my_array<int,3>{9,10,11}.data);
}

This works well, but requires one to add template argument to my_array, which are not deduced. With C++17 it is possible to automatically deduce type and size:

template <class T, std::size_t N>
struct my_array
{
    constexpr my_array(std::initializer_list<T> x)
    {
       std::size_t i = 0;
       for (auto val : x)
           data[i++] = val;
    }
    T data[N];
};
template <class ...T>
my_array(T...) -> my_array<typename std::common_type<T...>::type, sizeof...(T)>;

int main()
{
    print(my_array{9,10,11}.data);
}

For two dimensional arrays this is slightly more complicated:

template <class T, std::size_t N1, std::size_t N2>
struct my_array2d
{
    constexpr my_array2d(std::initializer_list<std::initializer_list<T> > x)
    {
        std::size_t i = 0;
        for (const auto & row : x) {
            int j=0;
            for (const auto & val: row) {
                data[i][j++] = val;
            }
            i++;
        }
    }
    T data[N1][N2];
};
int main()
{
    work(my_array2d<int, 3, 2>{{9,1},{10,2},{11,3}}.data);
}

I have given up on deduction guides for two dimensional arrays, but I believe they are possible.