David Golombek
6.035 pset #3
TA: Jeremy Brown
Problem 1

1a) 

expr ::= term
	| term '+' expr		{ Error if #1.attribs != #3.attribs }
	| term '*' expr		{ Error if #1.attribs != #3.attribs }
	| term '^' expr		{ Error if #1.attribs != #3.attribs != bool }

term ::= int
	| bool

1bi) true

1bii) 

PRODUCTION			SEMANTIC RULE
--------------------------------------------
S ::= E				print(E.val)
E ::= T				E.val = T.val
E ::= int + E			E_1.val = int.val + (bool->int E_2.val)
E ::= int * E			E_1.val = int.val * (bool->int E_2.val)
E ::= int ^ E			E_1.val = (int->bool int.val) XOR
					  (bool->int E_2.val)
E ::= bool + E			E_1.val = bool.val OR (int->bool E_2.val)
E ::= bool * E			E_1.val = bool.val AND (int->bool E_2.val)
E ::= bool ^ E			E_1.val = bool.val XOR (int->bool E_2.val)




David Golombek
6.035 pset #3
TA: Jeremy Brown
Problem 2

2a)
PRODUCTION			SEMANTIC RULE
--------------------------------------------
expr ::= term '*' term		expr.string = if term_1.is_simple
						then term_1.string
						else (term_1.string)
					      '*'
					      if term_2.is_simple
						then term_2.string
						else (term_2.string)
				expr.is_simple = false
expr ::= term '*' term		expr.string = if term_1.is_simple
						then term_1.string
						else (term_1.string)
					      '*'
					      if term_2.is_simple
						then term_2.string
						else (term_2.string)
				expr.is_simple = false
expr ::= term			expr.string = term.string
				expr.is_simple = term.is_simple

term ::= int_constant		term.string = int_constant
				term.is_simple = true
term ::= '(' expr ')'		term.string = expr.string
				term.is_simple = expr.is_simple

2b)
PRODUCTION			SEMANTIC RULE
--------------------------------------------
expr ::= term '+' term		expr.string = term_1.string '+' term_2.string
				expr.is_add = true
expr ::= term '*' term		expr.string = if term_1.is_add 
						then (term_1.string)
						else term_1.string
					      '*'
					      if term_2.is_add
						then (term_2.string)
						else term_2.string
				expr.is_add = false
expr ::= term			expr.string = term.string
				expr.is_add = false

term ::= int_constant		term.string = int_constant
term ::= '(' expr ')'		term.string = expr.string



David Golombek
6.035 pset #3
TA: Jeremy Brown
Problem 3

This problem is broken because there is no chaining on the 
one_box ::= SYMBOL rule, and thus no place to get the size 
passed from.


David Golombek
6.035 pset #3
TA: Jeremy Brown
Problem 4

4a)

        ----------			---------
	| animal |			| plant |
	----------			---------
      /           \
     /             \
-------------  -------------
| carnivore |  | herbivore |
-------------  -------------

4b)

This is legal, because carnivore inherits from animal, which has a
valid locate procedure.

Not legal, because plant$nutrition_value is a private method of plant,


4c) 

c = a;  Is not legal because c is a descendent of a and extends it.

a = h;  This is legal; h (herbivore) is typecast into an animal.