4. (19th "Hope Cup" Senior High School Competition Question) The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=3, a_{n+1}=-\frac{1}{a_{n}+1}$, then $a_{2008}$ equals
A. $-\frac{4}{3}$
B. $-\frac{1}{4}$
C. 3
D. -3
4. $\mathrm{C} \quad a_{1}=3, a_{2}=-\frac{1}{a_{1}+1}=-\frac{1}{4}, a_{3}=-\frac{1}{a_{2}+1}=-\frac{4}{3}, a_{4}=-\frac{1}{a_{3}+1}=3, \cdots$

It can be seen that the sequence $\left\{a_{n}\right\}$ has a period of 3. Therefore, from
$$
2008=669 \times 3+1
$$

we get $a_{2008}=a_{1}=3$, hence the answer is $\mathrm{C}$.