Is an affine scheme compact?

Question

This is probably a very stupid question, because it is very well known that the spectrum of a ring is compact (quasi-compact, if you want). But an affine scheme $(X,\mathcal{O}_X)$, if I am right, is just isomorphic, as a ringed space, to $(\operatorname{Spec}(A),\tilde{A})$ where $\tilde{A}$ is the canonical sheaf over $\operatorname{Spec}(A)$. So, the definition doesn't (to my understanding) imply that the topological space $X$ is homeomorphic to $\operatorname{Spec}(A)$.