As far as I understand correctly, to say that a decision problem $X$ is in P means that there exists a polynomial $c\cdot n^p$ such that as $n$ goes to $\infty$, the steps required to solve $X$ is bounded by $c\cdot n^p$.
But strictly speaking, this doesn't restrict the complexity for finite $n$ in any way. Strictly speaking if $X$ is in P, it could be that for all $n$ between $0$ and Ackermann function $A(300,300)$, the complexity of $X$ grows according to $e^n$, and then only from $A(300,300)$ onwards it starts being bounded by $c\cdot n^p$.
Am I right that if a problem is in the complexity class P this strictly speaking doesn't say anything about its complexity for finite $n$? Is there a way to talk about complexity for finite $n$?