The easiest solution would probably be to just let C do it. Per § 6.3.1.4 of the C11 spec:

When a finite value of real floating type is converted to an integer type other than _Bool, the fractional part is discarded (i.e., the value is truncated toward zero).

So, all you need to do is convert the value to an integer, than convert that back to a float:

float myFloor(float value) {
  return (float) (int) value;
}

If you need to handle negatives, you can easily do so:

float myFloor(float value) {
  float tmp = (float) (int) value;
  return (tmp != value) ? (tmp - 1.0f) : tmp;
}

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

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

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

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 $x$.