Consider two binary operators \(\text{‘} \uparrow \text{’}\) and \(\text{‘}…
2011
Consider two binary operators \(\text{‘} \uparrow \text{’}\) and \(\text{‘} \downarrow \text{’}\) with the precedence of operator \(\downarrow\) being lower than that of the operator \(\text{} \uparrow \text{}\). Operator \(\text{} \uparrow \text{}\) is right associative while operator \(\downarrow\) is left associative. Which one of the following represents the parse tree for expression \((7 \downarrow 3 \uparrow 4 \uparrow 3 \downarrow 2)\)
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Key rules:
↑ has higher precedence than ↓.
↑ is right-associative (group to the right).
↓ is left-associative (group to the left).
Apply the rules to the expression (7 ↓ 3 ↑ 4 ↑ 3 ↓ 2):
Because ↑ has higher precedence, group the ↑ operations first. Since ↑ is right-associative, 3 ↑ 4 ↑ 3 becomes 3 ↑ (4 ↑ 3).
After that grouping the expression becomes 7 ↓ (3 ↑ (4 ↑ 3)) ↓ 2.
Now apply ↓ left-to-right (left-associative), producing ((7 ↓ (3 ↑ (4 ↑ 3))) ↓ 2).
Final parse (parenthesized form): ((7 ↓ (3 ↑ (4 ↑ 3))) ↓ 2)
Common mistakes to avoid:
Grouping a ↓ under an ↑ (this violates precedence; ↑ must be grouped first).
Treating ↑ as left-associative (this yields (3 ↑ 4) ↑ 3 instead of 3 ↑ (4 ↑ 3)).
Applying ↓ before resolving the ↑ subexpressions (this also breaks precedence).
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