1. Given the function $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfies
$$
f(f(x)+f(y))=f(x)+y(\forall x, y \in \mathbf{R}) \text {. }
$$

then $f(2011)=$ $\qquad$ .
$-1.2011$.
Let $x=0$ in equation (1), we get
$$
f(f(0)+f(y))=f(0)+y,
$$

which shows that $f$ is surjective.
Let $x$ be the zero root $a$ of $f(x)$ in equation (1). Then $f(f(y))=y$.
Replace $x$ with $f(x)$ in equation (1) to get
$$
f(x+f(y))=x+y \text {. }
$$

Let $y=a$ in equation (2) to get $f(x)=x+a$.
Substitute into equation (1) to get
$$
f(x+a+y+a)=x+a+y \text {, }
$$

which simplifies to $x+y+3 a=x+y+a \Rightarrow a=0$.
Thus, $f(x)=x$.