How do I return an array reference from a const member function?
class State{
Chips (&arr() const)[6][7] { return arr_; }
Chips val() const { return val_; }
}
Invalid initialization of reference of type 'Chips (&)[6][7]' from expression of type 'const Chips [6][7]'
Thank you.
You got the syntax right, but if arr_ is an immediate member of the class (and it probably is), then you simply can't return a non-cost reference to this member. Inside the above arr method, member arr_ is seen as having const Chips[6][7] type. You can't use this type to initialize a reference of Chops (&)[6][7] type since it would break const-correctness. In order for the above to compile you'd need a const on the returned reference as well
...
const Chips (&arr() const)[6][7] { return arr_; }
...
But in any case, you'll be much better off with a typedef
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typedef Chips Chips67[6][7];
const Chips67 &arr() const { return arr_; }
...
const Chips (&arr() const)[6][7] { return err_; } works perfectly with g++-4.2, and I cannot see anything wrong with that syntax (besides the complexity to read it). What compiler are you using? (Anyway the typedef is almost a requirement for readability)You need to specify the function as returning a Chips pointer. So,
Chips* arr() const;
Is what you are looking for.
const Chips (*arr() const )[7] (and I am not even sure there). Note that only the outer array can/will decay automatically to a pointer, so if the type is int array[3][4][5], it is an array of 3 arrays of 4 arrays of 5 ints when it decays it will be a pointer to an array of 4 arrays of 5 ints. Also, note that the problem is that with the member method being const, the returned element must also be const --or the compiler will complain.