The first is the correct: you round "down" (i.e. the greatest integer LESS THAN $-0.8$).

In contrast, the ceiling function rounds "up" to the least integer GREATER THAN $-0.8 = 0$.

$$
\begin{align} \lfloor{-0.8}\rfloor & = -1\quad & \text{since}\;\; \color{blue}{\bf -1} \lt -0.8 \lt 0 \\ \\
\lceil {-0.8} \rceil & = 0\quad &\text{since} \;\; -1 \lt -0.8 \lt \color{blue}{\bf 0}
\end{align}$$

In general, we must have that $$\lfloor x \rfloor \leq x\leq \lceil x \rceil\quad \forall x \in \mathbb R$$

And so it follows that $$-1 = \lfloor -0.8 \rfloor \leq -0.8 \leq \lceil -0.8 \rceil = 0$$

K.Stm's suggestion is a nice, intuitive way to recall the relation between the floor and the ceiling of a real number $x$, especially when $x\lt 0$. Using the "number line" idea and plotting $-0.8$ with the two closest integers that "sandwich" $-0.8$ gives us:

$\qquad\qquad$

We see that the *floor* of $x= -0.8$ is the first integer immediately to the *left* of $-0.8,\;$ and the *ceiling* of $x= -0.8$ is the first integer immediately to the *right* of $-0.8$, and this strategy can be used, whatever the value of a real number $x$.