How to write a C++ conversion operator returning reference to array?
arrays, c++, implicit-conversion, operator-overloading
Solution
In fact you can, you almost had it with the last one:
(&operator float())[3];
As for the question of whether or not a typedef is ever necessary, I think it is from reading the comments on https://stackoverflow.com/a/6755760/2925619 (which answer is what helped me get the syntax for the above as well).
Edit:
Apparently, this syntax is incorrect and returning a reference to an array is forbidden as chris discovered for us. I guess you'll just have to settle for a `typedef`.
Problem
In C++ one can add implicit-conversion operators in a class or struct. For instance, 3D vector types usually include something like: ``` struct Vector { float x, y, z; operator float * () { return reinterpret_cast<float *>(this); } }; ``` to allow accessing the vector's elements with subscripts, passing to functions that want a pointer, etc. It occurred to me to wonder: can we instead write a conversion operator that returns a reference to array of float, instead of a pointer to float? (This is of purely academic interest. I don't know what benefits a reference-to-array would have, if any, over a simple pointer.) As a free function we can do this like: ``` float (&convert(Vector & v))[3] { return reinterpret_cast<float(&)[3]>(v); } Vector v; convert(v); ``` However, I haven't been able to find the right syntax to do this as a conversion operator. I've tried things like: ``` operator float(&)[3] () operator float(&())[3] float (&operator())[3] ``` and various other permutations, but I just get various syntax errors (g++ 4.8.1). Is it possible to write a conversion operator returning a reference to array, and if so, what is the syntax to do so?