为自定义类型EitherOr实现Eq类型类时遇语法错误求助

Hey there! Let's work through this parse error you're hitting with your EitherOr Eq instance. The error message you're seeing ("parse error on input '='") usually pops up when Haskell's parser can't properly distinguish between separate pattern-matching clauses—most often due to incorrect indentation or missing line breaks between definitions.

Let's Break Down the Problem in Your Code #

Your original code snippet has two main issues:

• You're trying to define multiple (==) clauses without proper line breaks and indentation, so the parser gets confused about where one clause ends and the next begins.
• The final Goodbye pattern is incomplete (you have Goodbye ... which is invalid syntax).

The Corrected Eq Instance #

Here's the proper way to write the instance, with all cases handled correctly:

-- First, make sure your data type is defined (just in case you haven't already)
data EitherOr a b = Hello a | Goodbye b

instance (Eq a, Eq b) => Eq (EitherOr a b) where
  (==) (Hello v) (Hello v') = v == v'
  (==) (Goodbye b) (Goodbye b') = b == b'
  (==) _ _ = False  -- Handle mismatched constructors (Hello vs Goodbye)

Key Fixes Explained: #

• Indentation & Line Breaks: Each (==) pattern-matching clause is on its own line, indented consistently (usually 4 spaces) under the where keyword. This tells the parser that each line is part of the Eq instance's definition.
• Complete Pattern Matching: We cover all possible cases:
  1. Two Hello constructors: compare their inner values with v == v'
  2. Two Goodbye constructors: compare their inner values with b == b'
  3. Any mismatched pair (one Hello, one Goodbye): return False immediately using the wildcard _ pattern.
• Valid Syntax: Every pattern is fully written out (no ... placeholders) and the right-hand side of each clause is a valid expression.

This should resolve the parse error and give you a fully functional Eq instance for your EitherOr type!

内容的提问来源于stack exchange，提问作者user2159471