$$
\begin{array}{l}
\text { [Solution] } \because a<0, \quad b<0, \quad \therefore \quad|b-c|=-(b-c) . \\
\therefore \quad|a|-|a+b|+|c-a|+|b-c| \\
\quad=-a-[-(a+b)]+(c-a)+[-(b-c)] \\
\quad=-a+a+b+c-a-b+c=2 c-a .
\end{array}
$$

Therefore, the correct choice is (C).
1.1 The real numbers $a, b, c$ are positioned on the number line as shown in the figure below:

Then the value of the algebraic expression $|a|-|a+b|+|c-a|+|b-c|$ is
(A) $-a$.
(B) $2 a-2 b$.
(C) $2 c-a$.
(D) $a$.
(E) 0.
(China Junior High School Mathematics Correspondence Competition, 1985)