jmc

On each face, there are $2$ diagonals like $x$. There are $6$ faces on a cube. Thus, there are $2\times 6=12$ diagonals that are "x-like". Every "y-like" diagonal must connect the bottom of the cube to the top of the cube. Thus, for each of the $4$ bottom vertices of the cube, there is a different "y-like" diagonal. So there are $4$ "y-like" diagonals. This gives a total of $12+4=16$ diagonals on the cube, which is answer $\boxed{16}$.