How do I initialize a member array with an initializer_list?

As far as I can tell, using list-initialization of the function argument of the constructor (8.5.4/1) should be legal and solves many of the issues of the above. However, GCC 4.5.1 on ideone.com fails to match the constructor and rejects it.

#include <array>

struct Foo
  {
    std::array< int, 2 > const data;

    Foo(std::array<int, 2> const& ini) // parameter type specifies size = 2
    : data( ini )
    {}
  };

Foo f( {1,3} ); // list-initialize function argument per 8.5.4/1

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If you really insist on initializer_list, you can use reinterpret_cast to turn the underlying array of the initializer_list into a C-style array.

Foo(std::initializer_list<int> ini) // pass without reference- or cv-qualification
: data( reinterpret_cast< std::array< int, 2 > const & >( * ini.begin() )

You can use a variadic template constructor instead of an initializer list constructor:

struct foo { 
    int x[2]; 
    template <typename... T> 
    foo(T... ts) : x{ts...} { // note the use of brace-init-list
    } 
};

int main() {
    foo f1(1,2);   // OK
    foo f2{1,2};   // Also OK
    foo f3(42);    // OK; x[1] zero-initialized
    foo f4(1,2,3); // Error: too many initializers
    foo f5(3.14);  // Error: narrowing conversion not allowed
    foo f6("foo"); // Error: no conversion from const char* to int
}

---

EDIT: If you can live without constness, another way would be to skip initialization and fill the array in the function body:

struct foo {
    int x[2]; // or std::array<int, 2> x;
    foo(std::initializer_list<int> il) {
       std::copy(il.begin(), il.end(), x);
       // or std::copy(il.begin(), il.end(), x.begin());
       // or x.fill(il.begin());
    }
}

This way, though, you lose the compile-time bounds checking that the former solution provides.