Rust Optimization Passes: Constant Folding and Dead Code Elimination

This challenge focuses on implementing fundamental optimization passes for a simple intermediate representation (IR) of code in Rust. You will build two distinct optimization techniques: constant folding and dead code elimination. These are crucial steps in compilers that reduce computation and simplify code, making programs faster and smaller.

Problem Description

Your task is to implement two optimization passes that operate on a simplified Abstract Syntax Tree (AST) or Intermediate Representation (IR) of arithmetic expressions.

The IR will be represented using an enum in Rust. The optimization passes should be applied sequentially. First, implement constant folding to evaluate constant subexpressions. Second, implement dead code elimination to remove unreachable code branches.

Key Requirements:

1. IR Definition: Define a Rust enum to represent the IR. This enum should support:

   • Integer literals (Int(i64)).
   • Binary operations (Add, Sub, Mul, Div) that take two operands.
   • Conditional branching (If { condition: Box<Expr>, then: Box<Block>, else: Box<Block> }).
   • No-operation (Nop).
   • A Block which is a Vec<Expr> representing a sequence of operations.

2. Constant Folding Pass:

   • This pass should recursively traverse the IR.
   • If a binary operation node has two integer literal operands, it should evaluate the operation and replace the node with the resulting integer literal.
   • Division by zero should be handled gracefully (e.g., by leaving the operation unevaluated or producing a specific error value, though for this challenge, we'll assume valid inputs or leave it as is).
   • The pass should return a new, optimized IR.

3. Dead Code Elimination Pass:

   • This pass should traverse the IR, specifically focusing on If statements.
   • If the condition of an If statement evaluates to a constant true or false (after constant folding), the else or then branch, respectively, should be removed.
   • If an If statement's condition is a constant true, the entire If statement can be replaced by its then block.
   • If an If statement's condition is a constant false, the entire If statement can be replaced by its else block (or Nop if both then and else are empty or become Nop).
   • After simplifying an If statement, the pass should recursively optimize the remaining branches.
   • The pass should also identify and replace standalone Nop expressions that are not part of a larger structure or where they are the sole element in a block.
   • The pass should return a new, optimized IR.

Expected Behavior:

• The optimization passes should be pure functions (or methods that return a new structure), meaning they don't modify the original IR in place.
• The passes should be applied in the order: Constant Folding, then Dead Code Elimination.
• The output IR should be functionally equivalent to the input IR.

Edge Cases to Consider:

• Division by zero (as mentioned, for simplicity, we can assume inputs that don't cause this, or leave the division unevaluated).
• Empty then or else blocks in If statements.
• Deeply nested If statements and expressions.
• Sequences of operations where some become Nop after optimization.

Examples

Example 1: Constant Folding

Input IR:
Add(Int(5), Int(3))

Output IR:
Int(8)

Explanation: The Add operation with two integer literals is evaluated to 8.

Example 2: Constant Folding with Nested Operations

Input IR:
Mul(Add(Int(2), Int(3)), Int(4))

Output IR:
Int(20)

Explanation: First, Add(Int(2), Int(3)) is folded to Int(5). Then, Mul(Int(5), Int(4)) is folded to Int(20).

Example 3: Dead Code Elimination (True Condition)

Input IR:
If {
    condition: Int(1), // Represents true
    then: Block(vec![Int(10), Int(20)]),
    else: Block(vec![Int(30)])
}

Output IR:
Block(vec![Int(10), Int(20)])

Explanation: The condition is a constant true. The 'else' block is discarded, and the 'If' statement is replaced by the 'then' block.

Example 4: Dead Code Elimination (False Condition)

Input IR:
If {
    condition: Int(0), // Represents false
    then: Block(vec![Int(10)]),
    else: Block(vec![Int(20), Int(30)])
}

Output IR:
Block(vec![Int(20), Int(30)])

Explanation: The condition is a constant false. The 'then' block is discarded, and the 'If' statement is replaced by the 'else' block.

Example 5: Combined Optimization

Input IR:
Block(vec![
    Add(Int(1), Int(1)), // Will be folded
    If {
        condition: Mul(Int(2), Int(0)), // Will be folded to 0 (false)
        then: Block(vec![Int(100)]),
        else: Block(vec![Int(200), Int(300)]) // This branch should be kept
    },
    Nop, // Should be removed if possible
    Int(5)
])

Output IR:
Block(vec![
    Int(2),
    Block(vec![Int(200), Int(300)])
])

Explanation:
1.  Constant folding: Add(Int(1), Int(1)) becomes Int(2). Mul(Int(2), Int(0)) becomes Int(0).
2.  Dead code elimination: The If statement's condition is Int(0) (false). The 'then' block is discarded, and the 'If' becomes its 'else' block: Block(vec![Int(200), Int(300)]).
3.  Nop removal: The standalone Nop is removed from the outer block.
4.  The final output is a Block containing the folded Add result and the optimized If statement's else branch.

Constraints

• Integer literals will be within the range of i64.
• The input IR will not contain any undefined behavior that would prevent constant folding (e.g., division by zero will not occur in valid test cases, or if it does, it's acceptable for the operation to remain unevaluated).
• The depth of recursion for optimization passes should not exceed reasonable limits for typical program structures. Aim for O(N) complexity where N is the number of nodes in the IR.

Notes

• Consider using a recursive approach for both optimization passes.
• For If statements, you'll need to handle the condition, then block, and else block. The condition might itself be a complex expression that needs to be evaluated for constant folding.
• When a branch of an If statement is removed, ensure that the remaining branch is also recursively optimized.
• The Nop node is a placeholder. After other optimizations, some If branches or entire If statements might become effectively empty or redundant, which could be simplified further by treating them as Nops that can then be eliminated by the dead code pass.
• Think about how to represent the IR. A common pattern is an enum where variants hold boxed children.

This challenge provides a solid foundation for understanding compiler optimization techniques. Good luck!