Challenge 3: Does Either Language's main Get a Free Pass? — Possible Solution
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// Java's `public static void main(String[] args)` genuinely gets a
// free pass from Java's own type system -- there is no mechanism in
// Java's types at all that distinguishes "a method that only computes
// a value" from "a method that reads files, opens sockets, and prints
// to the console." main is free to do absolutely anything, and
// nothing about its signature says so one way or the other -- the
// SAME is true of every other Java method too, which is exactly why
// there's nothing special about main specifically getting a pass: no
// Java method is ever tracked for side effects, main included.
//
// Haskell's `main :: IO ()`, by contrast, does NOT get a free pass --
// and that's the entire point of this chapter. main's type explicitly
// declares that it may perform IO, exactly the same way any other
// IO-performing function's type would. There is no special, untracked
// "root" function in Haskell that's allowed to silently do IO without
// its type saying so. If main's own type were declared as something
// that claimed to be pure (which isn't actually possible for a real
// program that needs to interact with the world, but hypothetically),
// GHC would refuse to compile a main that tried to call putStrLn or
// getLine inside it, exactly the same rejection any other pure
// function calling IO code would get.
//
// The real difference isn't that Haskell has SPECIAL rules for main --
// it's that Haskell has NO exceptions to its one general rule (every
// side-effecting computation is tracked in its type), while Java has
// no such rule to begin with, for main or for anything else.

WHY THIS WORKS AS AN ANSWER
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This correctly identifies that Java's main isn't specially exempt --
NOTHING in Java is tracked for side effects, main included -- while
Haskell's main is genuinely bound by the same IO-tracking rule as
every other function, matching the chapter's own "no trusted root"
claim precisely.