Consider the following grammar along with translation rules. \[ S \rightarrow S_1 \# T \,\,\,\,\,\,\{S_{\centerdot\text{val}} = S_{1\centerdot\text{val}}* T_{\centerdot\text{val}}\} \] \[ S \rightarrow T \,\,\,\,\,\,\{S_{\centerdot\text{val}} = T_{\centerdot\text{val}}\} \] \[ T \rightarrow T_1 \% R \,\,\,\,\,\, \{T_{\centerdot\text{val}} = T_{1\centerdot\text{val}} \div R_{\centerdot\text{val}}\} \] \[ T \rightarrow R \,\,\,\,\,\,\{T_{\centerdot\text{val}} = R_{\centerdot\text{val}}\} \] \[ R \rightarrow \text{id} \,\,\,\,\,\,\{R_{\centerdot\text{val}} = \text{id}_{\centerdot\text{val}}\} \] Here \(\#\) and % are operators and id is a token that represents an integer and \( \text{id}_{\text{val}} \) represents the corresponding integer value. The set of non-terminals is {S, T, R, P}, and a subscripted non-terminal indicates an instance of the non-terminal.
Using this translation scheme, the computed value of} \( S_{\text{val}} \) for root of the parse tree for the expression \(20\#10%5\#8%2\#2 \) is
Given Expression:
$20 \# 10 \% 5 \# 8 \% 2 \# 2$
Step 1: Identify Operator Precedence
In the provided grammar logic, the modulus/division operator (%) is nested deeper in the derivation rules than the multiplication/concatenation operator (#). This means all $\%$ operations must be evaluated first to form the terms for the $\#$ operations.
Step 2: Evaluate High-Precedence Terms (%)
We identify the sub-expressions separated by the $\#$ operator and calculate the modulus values:
Step 3: Evaluate Final Expression (#)
Now, substitute the evaluated terms back into the original expression structure, which now consists only of the $\#$ operator:
$S_{val} = 20 \# 0 \# 0 \# 2$
Applying the multiplication rule for $\#$:$$S_{val} = 20 \times 0 \times 0 \times 2$$$$S_{val} = \mathbf{80}$$
Final Answer:
The final value of the expression is: 80