4. (12th "Hope Cup" Senior High School Competition) The function $f(x)$ defined on the set of real numbers satisfies $f(x-1)=$ $\frac{1+f(x+1)}{1-f(x+1)}$, then the value of $f(1) \cdot f(2) \cdot f(3) \cdot \cdots \cdot f(2008)+2008$ is
A. 2007
B. 2008
C. 2009
D. 0
4. C H Given $f(x)=\frac{1+f(x+2)}{1-f(x+2)}$,

So $f(x+1)=\frac{1+f(x+3)}{1-f(x+3)}$.
From $f(x-1)=\frac{1+f(x+1)}{1-f(x+1)}$ we get
$f(x+1)=\frac{f(x-1)-1}{f(x-1)+1}$.
Therefore, $\frac{1+f(x+3)}{1-f(x+3)}=\frac{f(x-1)-1}{f(x-1)+1}$.
So $f(x-1) \cdot f(x+3)=-1$.
We can infer $f(x+3) \cdot f(x+7)=-1$.
Thus, $f(x-1)=f(x+7)$.
That is, $f(x)=f(x+8)$.
Hence, $f(x)$ is a periodic function with a period of 8.
Since $f(1) \cdot f(5)=f(2) \cdot f(6)=f(3) \cdot f(7)=f(4) \cdot f(8)=-1$.
Therefore, $f(1) \cdot f(2) \cdot f(3) \cdot f(4) \cdot f(5) \cdot f(6) \cdot f(7) \cdot f(8)=1$.
So the original expression $=1+2008=2009$. Hence, the answer is C.