Chapter 4, Problem 12E

Explanation of Solution

Creating a grammar:

• One of the potential criticism of the right solution for the pervious Exercise 11, the problem is that the values of in the internal nodes of the parse tree do not reflect the value in context.
• Here, an alternative solution that addresses in the Exercise 11.
• In this solution, the val of an internal node is the sum of the vals of its children.
• The following grammar shows an underlying CFG, and non-L-attributed flow.

            $\begin{array}{l}\text{C }\to \text{ L\_digits }\text{. R\_digits}\\ \Rightarrow \text{R\_digits}\text{.pos := -1}\\ \Rightarrow \text{C}\text{.val := L\_digits}\text{.val + R\_digits}\text{.val}\\ \text{L\_digits }\to \text{ digit more\_L\_digits}\end{array}$

            $\begin{array}{l}\Rightarrow \text{digit}\text{.pos := more\_L\_digits}\text{.len}\\ \Rightarrow \text{L\_digits}\text{.len := more L digits}\text{.len + 1}\\ \Rightarrow \text{L\_digits}\text{.val := digit}\text{.val + more\_L\_digits}\text{.val}\\ \text{more\_L\_digits }\to \text{L\_digits}\end{array}$

            $\begin{array}{l}\text{}\Rightarrow \text{}\text{}\text{}\text{}\text{}\text{more\_L\_digits}\text{.len := L\_digits}\text{.len}\\ \Rightarrow \text{more\_L\_digits}\text{.val := L\_digits}\text{.val}\\ \text{more\_L\_digits }\to \in \\ \Rightarrow \text{more\_L\_digits}\text{.len := 0}\end{array}$

            $\begin{array}{l}\Rightarrow \text{more\_L\_digits}\text{.val := 0}\\ \text{R\_digits }\to \text{ digit more\_R\_digits}\\ \Rightarrow \text{digit}\text{.pos := R\_digits}\text{.pos}\\ \Rightarrow \text{more\_R\_digits}\text{.pos := R\_digits}\text{.pos - 1}\end{array}$

            $\begin{array}{l}\Rightarrow \text{R\_digits}\text{.val := digit}\text{.val + more\_R\_digits}\text{.val}\\ \text{more\_R\_digits }\to \text{ R\_digits}\\ \Rightarrow \text{R\_digits}\text{.pos := more\_R\_digits}\text{.pos}\\ \Rightarrow \text{more\_R\_digits}\text{.val := R\_digits}\text{.val}\end{array}$

            $\begin{array}{l}\text{more\_R\_digits }\to \in \\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\Rightarrow \text{more\_R\_digits}\text{.val := 0}\\ \text{digit }\to \text{ 0}\\ \Rightarrow \text{digit}\text{.val := 0 }\times {\text{ 10}}^{\text{digit}\text{.pos}}\end{array}$

            $\begin{array}{l}\text{digit }\to \text{ 1}\\ \Rightarrow \text{digit}\text{.val := 1 }\times {\text{ 10}}^{\text{digit}\text{.pos}}\\ \text{digit }\to \text{ 2}\\ \Rightarrow \text{digit}\text{.val := 2 }\times {\text{ 10}}^{\text{digit}\text{.pos}}\end{array}$

            $\begin{array}{l}\text{digit }\to \text{ 3}\\ \Rightarrow \text{digit}\text{.val := 3 }\times {\text{ 10}}^{\text{digit}\text{.pos}}\\ \text{digit }\to \text{ 4}\\ \Rightarrow \text{digit}\text{.val := 4 }\times {\text{ 10}}^{\text{digit}\text{.pos}}\end{array}$

            $\begin{array}{l}\text{digit }\to \text{ 5}\\ \Rightarrow \text{digit}\text{.val := 5 }\times {\text{ 10}}^{\text{digit}\text{.pos}}\\ \text{digit }\to \text{ 6}\\ \Rightarrow \text{digit}\text{.val := 6 }\times {\text{ 10}}^{\text{digit}\text{.pos}}\end{array}$

            $\begin{array}{l}\text{digit }\to \text{ 7}\\ \Rightarrow \text{digit}\text{.val := 7 }\times {\text{ 10}}^{\text{digit}\text{.pos}}\\ \text{digit }\to \text{ 8}\\ \Rightarrow \text{digit}\text{.val := 8 }\times {\text{ 10}}^{\text{digit}\text{.pos}}\end{array}$

            $\begin{array}{l}\text{digit }\to \text{ 9}\\ \Rightarrow \text{digit}\text{.val := 9 }\times {\text{ 10}}^{\text{digit}\text{.pos}}\end{array}$

Parse Tree for the grammar:

• The double boxes on digit, R_digit and more_R_digit represents the pos (left) and val (right).
• The boxes on L_digits and more_L_digits represent len (left) and val (right).
• The single box represent for G is val.