CODEWITHSUNDEEP
prolog
InquisitiveCoder
InquisitiveCoder

Reputation: 171

Prolog lexer can't find the error

I have a lexer that I am trying to build. I just want to know why it doesn't create the second list.

%TokenList=[int, add, '(', int, a, ',', int, b, ')']. 


lexer([H|TokenList],Output,[Output|X2]):-
   lexer_test(H,Output),lexer(TokenList,Output,X2).

lexer_test(H,Output):-
   (
    H = '\'int\'' -> Output = 'TYPE_INT'
    ;(H = '\'bool\'' -> Output='TYPE_BOOL'
    ;(H = '\',\'' -> Output='COMMA'
    ;(H = '\'=\'' -> Output='ASSIGN'
    ;(H = '\'let\'' -> Output='LET'
    ;(H = '\'in\'' -> Output='LET_IN'
    ;(H = '\'if\'' -> Output='COND_IF'
    ;(H = '\'then\'' -> Output='COND_THEN'
    ;(H = '\'else\'' -> Output='COND_ELSE'
    ;(H = '\'==\'' -> Output='LOGIC_EQ'
    ;(H = '\'!=\'' -> Output='LOGIC_NOT_EQ'
    ;(H = '\'>\'' -> Output='LOGIC_GT'
    ;(H = '\'>=\'' -> Output='LOGIC_GTEQ'
    ;(H = '\'+\'' -> Output='ARITH_ADD'
    ;(H = '\'-\'' -> Output='ARITH_SUB'
    ;(H = '\'(\'' -> Output='OPEN_P'
    ;(H = '\')\'' -> Output='CLOSE_P'
    ;(H = '\'0\'' -> Output='INTEGER'
     ;(H = '\'1\'' -> Output='INTEGER'
     ;(H = '\'2\'' -> Output='INTEGER'
     ;(H = '\'3\'' -> Output='INTEGER'
     ;(H = '\'4\'' -> Output='INTEGER'
     ;(H = '\'5\'' -> Output='INTEGER'
     ;(H = '\'6\'' -> Output='INTEGER'
     ;(H = '\'7\'' -> Output='INTEGER'
     ;(H = '\'8\'' -> Output='INTEGER'
     ;(H = '\'9\'' -> Output='INTEGER' 
    ;(Output='IDENTIFIER'
    )))))))))))))))))))))))))))).

Don't worry about the second function lexer_test. I can fix that. But I really don't know why I don't get an output list. And if I can make lexer/2 instead of lexer/3 that would be great. And if you could suggest some good reads to learn prolog please do let me know. Thanks.

Upvotes: 0

Views: 57

Answers (1)

SND
SND

Reputation: 1557

At the moment, you do not specify a termination condition:

lexer([H|TokenList],Output,[Output|X2]):-
   lexer_test(H,Output),lexer(TokenList,Output,X2).

Lexer keeps calling itself (recursion), but you have no other clause specified, and thus Prolog will end up in an endless loop. The condition you are looking for is:

lexer([],_,[]).

as stop condition. This states the condition in which Tokenlist becomes empty.

A good place to start learning prolog (imho) would be: http://www.learnprolognow.org/

Upvotes: 1

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