Grupo 06 - Dicionários

[enzzoMs]

This Pull Request contains the necessary changes for the implementation of the dictionary data type. Dictionaries are data structures that allow storing key-value pairs, where each key is unique and associated with a specific value.

-> Group Members:

Gabriel Xisto Barros: 211055503
Enzzo Morais de Olinda: 211042720
Davi Mansur Costa: 211042701

-> Main Contributions:

Added support for:

• Dictionary expressions.
  Example: dict = { x: 10, y: 20 }

• Dictionary key access.
  Example: dict.x

• Dictionary key assignment.
  Example: dict.x = 15

• Dictionary delete key statement.
  Example: del dict.x

• Dictionary merge expression. The merge expression creates a new dictionary that combines the fields of the first dictionary with those of the second, overwriting duplicate key values.
  Example: new_dict = dict with { z = 15 }

• Membership operators in and not in for dictionaries.
  Example: x in dict, x not in dict

To implement these features, modifications were made to the interpreter, type checker, AST, and parser modules. The implementation was validated through 25 tests.

-> Limitations/Issues:

• Only identifiers can be used as dictionary keys. It is not possible to use strings, integers, or other types as keys directly.
  Example: dict = { "key": 10, 42: 20 } -> Error. Only identifiers are allowed as keys.

• Dictionary expressions can include duplicate keys, which may lead to unexpected behavior.
  Example: dict = { x: 10, x: 20 }

• Key access and assignment can only be performed on variables. It is not possible to access or modify keys from an expression that returns a dictionary.
  Examples that are NOT allowed:
  dict.key.key2 -> Even if dict.key returns a valid dictionary, chained access is not allowed.
  fun().key -> Even if the function returns a dictionary, its keys cannot be accessed directly.
  { x: 10 }.x -> Dictionary expressions cannot be accessed directly.

• It is not possible to add a key directly to an existing dictionary. The only way to add a key is through the with operator, but this creates a new dictionary without modifying the original.