4. The area of the figure enclosed by the curve determined by the equation $|x-1|+|y-1|=1$ is
(A) 1;
(B) 2;
(C) $\pi$;
(D) 4.
4. (B)

For expressions containing absolute values, discuss cases to remove the absolute value symbols, resulting in four line segments.
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ x - 1 \geqslant 0 } \\
{ y - 1 \geqslant 0 } \\
{ x + y \geqslant 3 , }
\end{array} \quad \left\{\begin{array}{l}
x-1 \leqslant 0 \\
y-1 \leqslant 0 \\
x+y=1,
\end{array}\right.\right. \\
\left\{\begin{array} { l } 
{ x - 1 \leqslant 0 } \\
{ y - 1 \geqslant 0 } \\
{ y = x + 1 , }
\end{array} \quad \left\{\begin{array}{l}
x-1 \geqslant 0 \\
y-1 \leqslant 0 \\
y=x-1 .
\end{array}\right.\right.
\end{array}
$$

The curves determined by them are
a square with vertices at $(1,0),(2,1),(1,2),(0,1)$, with a side length of $\sqrt{2}$, thus the area is 2.