Exercise 2: Reading One Global 300 Times, Without Deduplication — Possible Solution
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THE SETUP
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    class ChunkNoDedup:
        def add_constant(self, value):
            self.constants.append(value)          # no "if value in self.constants" check
            return len(self.constants) - 1

    # var g = 1;  then  g;  three hundred times
    program = [VarStmt('g', Literal(1.0), line=1)]
    for _ in range(300):
        program.append(ExpressionStmt(Variable('g'), line=1))

RESULT
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    ValueError: byte must be in range(0, 256)

WHY THIS WORKS AS AN ANSWER
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The program only ever refers to ONE variable name, 'g' -- but every
single one of the 300 reads calls visit_variable, which calls
self.chunk.add_constant(node.name) independently, with no memory of
whether that exact string was already added a moment ago. Without
the dedup check, add_constant() blindly appends 'g' as a brand new
constant-pool entry every time it's called, so after 256 reads the
constant pool already holds 256 separate (but identical-valued)
entries, and the 257th read's own OP_GET_GLOBAL operand byte can no
longer be represented in a single byte.

This is exactly the ceiling Chapter 2 documented for numeric
literals, showing up again for exactly the same structural reason,
but triggered by something that looks completely harmless at the
Wisp source level -- reading the same variable repeatedly is about
as ordinary as Wisp code gets. This is precisely why this chapter's
own Compiler reuses Chapter 2 Exercise 2's dedup fix for
add_constant() rather than treating it as optional polish: without
it, ANY moderately long loop that reads a global more than 256 times
would fail to compile at all, which would make this VM unusable for
almost any real program with a loop in it.