4. Given an odd function $f(x)$ defined on $\mathbf{R}$ whose graph is symmetric about the line $x=2$, when $0<x \leq 2$, $f(x)=x+1$, then $f(-100)+f(-101)=$ $\qquad$ .
Answer 2.
Analysis: Given that $f(x)$ is an odd function and its graph is symmetric about the line $x=2$, we know that $f(-x)=-f(x)$, and $f(2-x)=f(2+x)$, so
$$
f(x+4)=f(-x)=-f(x), \quad f(x+8)=-f(x+4)=f(x)
$$
$f(x)$ is a periodic function with a period of 8.
Also, $f(3)=f(1)=2, f(4)=f(0)=0$, so $f(-100)+f(-101)=f(4)+f(3)=0+2=2$.