How do I initialize a member array with an initializer_list?

Question

I'm getting up to speed with C++0x, and testing things out with g++ 4.6

I just tried the following code, thinking it would work, but it doesn't compile. I get the error:

incompatible types in assignment of ‘std::initializer_list<const int>’ to ‘const int [2]’

struct Foo   {     int const data[2];      Foo(std::initializer_list<int const>& ini)     : data(ini)     {}   };  Foo f = {1,3}; 

Answer 1

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.