Generic function using std::span doesn't compile

Question

I wanted to make an isIn function that takes an std::span.

This is my attempt:

#include <span>

template <typename T1, typename T2>
bool isIn(const T1& x, std::span<const T2> v)
{
    for (const T2& e : v)
        if (e == x)
            return true;
    return false;
}

// this one would work, but I want my function to be generic
/*bool isIn(int x, std::span<const int> v)
{
    for (int e : v)
        if (e == x)
            return true;
    return false;
}*/

int main()
{
    const int v[] = {1, 2, 3, 4};
    isIn(2, v); // I want this, but doesn't compile
    //isIn(2, std::span<const int>(v)); // this works fine
}

As you can see, I can get around by doing this casting:

isIn(2, std::span<const int>(v));

But that's quite verbose, and I would like to do something like this:

isIn(2, v);

Is there any way that can be achived?

https://godbolt.org/z/czTs83

Answer 1

There are no conversion/promotion for template deduction,

so const int (&)[4] cannot be deduced as std::span<const int /*, 4*/>.

You might still provide overload to do the conversion yourself (care to avoid infinite recursive call):

template <typename T1, typename T2, std::size_t N>
bool isIn(const T1& x, std::span<const T2, N> v)
{
    // return std::find(std::begin(v), std::end(v), x) != std::end(v);
    for (const T2& e : v) {
        if (e == x) {
            return true;
        }
    }
    return false;
}

template <typename T, typename C>
bool isIn(const T& x, const C&c)
{
    return isIn(x, std::span(c)); // Use CTAD for the conversion.
}

But std::span is not needed here:

template <typename T, typename C>
bool isIn(const T& x, const C& c)
{
    // return std::find(std::begin(c), std::end(c), x) != std::end(c);
    for (const auto& e : c) {
        if (e == x) {
            return true;
        }
    }
    return false;
}

Answer 2

This implementation covers all my use cases. I though it might be useful to someone else:

#include <stdio.h>
#include <initializer_list>
#include <string_view>

template <typename T1, typename T2>
bool isIn(const T1& x, const T2 &v)
{ 
    printf("generic\n");
    for (auto& e : v)
        if (e == x)
            return true;
    return false;
}

bool isIn(char x, const char* v)
{
    // NOTE: we have this specialization because we don't want to consider the null terminator
    printf("str\n");
    return isIn(x, std::string_view(v));
}

template <typename T1, typename T2>
bool isIn(const T1& x, std::initializer_list<T2> v)
{ 
    printf("initializer_list\n");
    for (auto& e : v)
        if (e == x)
            return true;
    return false;
}

int main()
{
    const int v[] = {1, 2, 3, 4};
    isIn(2, v); // generic
    isIn(2, {1, 2, 3, 4}); // initializer_list
    isIn('l', "blabla"); // str
}

https://godbolt.org/z/d3nW73

Answer 3

Here is a c++20 version of your code.

First we start off with two concepts; is_span and spannable:

template<class T>
concept is_span = requires(T& a) {
    { std::span(a) } -> std::same_as<T>;
};
template<class T>
concept spannable = requires(T& a) {
    { std::span(a) };
} && !is_span<T>;

Something spannable can be deduced into a span without being one.

Then we write two overloads:

constexpr bool isIn(const auto& x, is_span auto const& v)
{
    for (const auto& e : v)
        if (e == x)
            return true;
    return false;
}

constexpr bool isIn(const auto& x, spannable auto const& v)
{
    return isIn(x, std::span(v));
}

using the new syntax.

We then add one more overload:

template<std::size_t N>
constexpr bool isIn(const auto& x, auto const(& v)[N])
{
    return isIn(x, std::span(v));
}

which permits this tasty syntax:

static_assert( isIn( 7, {1,2,3,4,5,6,7} ));

Live example

Now all you have to do is make it infix.

That "sadly" means

static_assert( isIn('\0', "hello") );

is true, as "hello" is an array containing a '\0' at the end.

template<class T>
constexpr bool isIn(const auto& x, std::initializer_list<T> il)
{
    return isIn(x, std::span(il));
}
template<std::size_t N>
constexpr bool isIn(char x, char const(& v)[N])
{
    return isIn(x, std::span(v, v+N-1));
}

Live example.

Answer 4

There is no need to use std::span in generic functions like this one:

template <typename T1, typename T2>
bool isIn(T1&& x, T2 &&v) // always use universal references
{ 
    for (auto&& e : v)
        if (e == x)
            return true;
    return false;
}

int main()
{
    const int v[] = {1, 2, 3, 4};
    isIn(2, v); // T2 is const &int[4]
    isIn(2, std::span<const int>{v}); // T2 is span<const int>
    isIn(2, std::list<int>(std::begin(v),std::end(v)) ); // T2 is list<int>
       // but list cannot be a span!

    isIn('l',"blablalba"); //works too, T2 is const &char[9]

    std::string_view s = "blablalba";
    isIn('l',s); // T2 is std::string_view&

}

This way it will work for any type that have std::begin and std::end defined.