Implementing Peephole Optimization for Go Bytecode

This challenge involves implementing a simplified peephole optimizer for Go's intermediate representation (IR). Peephole optimization is a technique that examines a small, fixed window of instructions (the "peephole") and replaces inefficient instruction sequences with equivalent, but more efficient ones. This can lead to faster and smaller compiled Go programs.

Problem Description

Your task is to write a Go program that takes a sequence of hypothetical Go bytecode instructions as input and applies a set of predefined peephole optimization rules to reduce redundant or inefficient patterns. You will need to parse the input bytecode, identify matching patterns within a sliding window, and replace them with optimized sequences.

Key Requirements:

1. Instruction Representation: Define a suitable Go struct or interface to represent individual bytecode instructions. Each instruction should have an Opcode (e.g., ADD, MOV, JMP) and potentially operands.
2. Peephole Window: Implement a mechanism to slide a fixed-size window (e.g., 2 or 3 instructions) over the instruction sequence.
3. Optimization Rules: Define and implement at least three distinct peephole optimization rules. Each rule should identify a specific pattern of instructions and replace it with a more efficient equivalent.
4. In-Place or New Sequence: The optimizer can either modify the instruction sequence in-place or generate a new, optimized sequence.
5. Output: The program should output the optimized sequence of bytecode instructions.

Expected Behavior:

The program should process the input bytecode, systematically applying optimization rules wherever they match within the peephole window. The optimization should be greedy, meaning if a pattern matches and is replaced, the optimizer should re-evaluate the window from the beginning of the replaced section to catch cascading optimizations.

Edge Cases to Consider:

• Instruction sequences shorter than the peephole window.
• Patterns matching at the very beginning or end of the instruction sequence.
• Optimizations that might change control flow (e.g., dead code elimination after jumps).

Examples

Example 1: Redundant Move

Input:
[
  {Opcode: "MOV", Operands: ["a", "b"]}, // Move value of b to a
  {Opcode: "MOV", Operands: ["a", "c"]}  // Move value of c to a
]

Output:
[
  {Opcode: "MOV", Operands: ["a", "c"]}
]

Explanation: The first MOV instruction is redundant because the value of 'b' is immediately overwritten by 'c' in the second instruction. The second instruction effectively achieves the desired outcome for 'a'.

Example 2: Dead Code Elimination (Conditional Jump)

Assume x == 0 is a condition that evaluates to false.

Input:
[
  {Opcode: "LOAD_CONST", Operands: [0]},       // Load constant 0
  {Opcode: "STORE_VAR", Operands: ["x"]},     // Store 0 in variable x
  {Opcode: "JUMP_IF_FALSE", Operands: ["label1"]}, // If x is false (non-zero), jump
  {Opcode: "LOAD_CONST", Operands: [1]},       // This instruction will not be executed if x is 0
  {Opcode: "ADD", Operands: ["y", "z"]},       // This instruction will not be executed if x is 0
  {Opcode: "label1"}                           // Label to jump to
]

Output:
[
  {Opcode: "LOAD_CONST", Operands: [0]},
  {Opcode: "STORE_VAR", Operands: ["x"]},
  {Opcode: "label1"}
]

Explanation: Since "x" is explicitly set to 0, the `JUMP_IF_FALSE` (assuming it jumps if the condition is *false*, i.e., x is non-zero) will always take the jump if x is 0. The instructions between the jump and the label are unreachable and can be eliminated. (Note: The exact semantics of `JUMP_IF_FALSE` might vary; this example assumes a common interpretation for illustration).

Example 3: Constant Folding

Input:
[
  {Opcode: "LOAD_CONST", Operands: [5]},
  {Opcode: "LOAD_CONST", Operands: [3]},
  {Opcode: "ADD"}
]

Output:
[
  {Opcode: "LOAD_CONST", Operands: [8]}
]

Explanation: The two LOAD_CONST instructions load constants 5 and 3. The ADD instruction then adds them. This entire sequence can be replaced by a single instruction that loads the pre-computed result, 8.

Constraints

• The input bytecode will be a slice of instruction objects.
• Opcodes will be represented as strings. Operands can be strings (variable names) or integers (constants).
• The peephole window size will be fixed at 3 instructions for the purpose of this challenge.
• The number of input instructions will not exceed 1000.
• The optimizer should aim for correctness over extreme performance.

Notes

• You'll need to define your own Instruction struct and a set of opcodes. For simplicity, you can hardcode the optimization rules.
• Consider how to handle operand types and matching.
• Think about how to manage the instruction list when replacements occur.
• The provided examples use simplified representations. Real Go bytecode is more complex, but this challenge abstracts it for focus.
• Some rules might require looking ahead or behind the immediate neighbors within the peephole window.