Regular of language of all words of length 3 Consider the language $$L = \{ x \in \{0,1\}^* \mid |x| = 3 \}.$$
I think the above language is regular. A DFA can be used to determine the above language.

Am I correct? Is the above language regular?

If this language $L$ is regular, then it should satisfy the pumping lemma. Then there exist $w = xyz$, where $xy^nz \in L$ for all $n \ge 0$.
But on the other hand, if we pump more letters then the resulting string will not be in the language. The language $L$ only contains words of length 3.
The pumping lemma states that for every regular language there exists an integer $p$, such that string $w$ of length at least $p$ can be written as $w = xyz$ and $y$ can be pumped.
Here are my doubts.

Is this language $L$ regular?
If so, does it satisfy the pumping lemma?
The pumping lemma states that every regular language has a pumping length $p \ge 1$. Does this language not have one?
Every finite language is regular. If $L$ is a finite language and $p$ is larger than the length of all words in $L$, then $L$ satisfies the pumping lemma with the constant $p$. Indeed, every word in $L$ of length at least $p$ can be pumped (vacuously).