Let the constant term of Charles's quadratic be $ c $, and the constant term of Eric's quadratic be $ d $. Then Charles's discriminant is $ (4)^2-4(1)(c)=16-4c $, and Eric's discriminant is $ (4)^2-4(1)(d)=16-4d $. We're given that $$\frac{\text{Discriminant}_{\text{Eric}}}{\text{Discriminant}_{\text{Charles}}}=\frac{\text{Constant}_{\text{Charles}}}{\text{Constant}_{\text{Eric}}},$$or $ \frac{16-4d}{16-4c}=\frac{c}{d} $. Cross multiplying gives \begin{align*} d(16-4d)&=c(16-4c)\quad\Rightarrow\\ 16d-4d^2&=16c-4c^2\quad\Rightarrow\\ 4c^2-4d^2&=16c-16d\quad\Rightarrow\\ 4(c+d)(c-d)&=16(c-d).\end{align*}Since $ c\neq d $, we know that $ c-d\neq 0 $, so we can cancel this term to find \begin{align*} 4(c+d)&=16\quad\Rightarrow\\ c+d&=4.\end{align*}Thus the sum of Eric's and Charles's constant terms is $ \boxed{4} $.