Exercise to understand Lisp cons/car/cdr, by implementing in JS

cons.html

<script src="cons.js"></script>

cons.js

//cons, car and cdr
(function( global ) { 

  var Cons = {
    
    cons: function() {      
      var ret = [], 
          args = [].slice.call( arguments, 0 ), 
          len = args.length, 
          idx, arg;

      for ( idx = 0; idx < len; idx++ ) {

        var arg = args[ idx ];

        ret = ret.concat( arg );
      }

      return ret;
    }, 
    
    //  These need to be broken out and abstracted for DRYer code
    car: function() {
    
      var car = this.cons.apply( null, [].slice.call( arguments, 0 )  );
      
      return car.splice( 0, 1 );
    }, 
    cdr: function() {
      
      var cdr = this.cons.apply( null, [].slice.call( arguments, 0 ) );
    
      return cdr.splice( 1, cdr.length );
    }
  }; 

  
  global.cons = Cons.cons;
  global.car = Cons.car;
  global.cdr = Cons.cdr;
  

  
})( this );  


//console.log( 
//  Actual, 
//  Expected
//);



console.log(
  "cons", 
  cons( [ "foo", "bar", "baz", "qux" ] ), 
  [ "foo", "bar", "baz", "qux" ]
);

console.log( 
  "cons", 
  cons( [ "foo", "bar", "baz", "qux" ] , [ "alpha", "beta" ] ), 
  [ "foo", "bar", "baz", "qux", "alpha", "beta" ]
);

console.log( 
  "cons", 
  cons( ["foo"], cons( ["bar", "baz", "qux"], cons( ["alpha"] ), cons(["beta"]) ) ), 
  
  [ "foo", "bar", "baz", "qux", "alpha", "beta" ]
);



console.log( 
  "car",
  car( [ "foo", "bar", "baz", "qux" ] ), 
  [ "foo" ]
);

console.log(
  "cdr", 
  cdr( [ "foo", "bar", "baz", "qux" ] ), 
  [ "bar", "baz", "qux" ]
);


console.log(
  "car(cdr())", 
  car( cdr( [ "foo", "bar", "baz", "qux" ] ) ), 
  [ "bar" ]
);

//  Composites

function cadr( x ) {
  return car( cdr( x ) );
}

console.log( 
  "cadr", 
  cadr( [ "foo", "bar", "baz", "qux" ] ), 
  [ "bar" ]
);

function caar( x ) {
  return car( car( x ) );
}


console.log(
  "caar", 
  caar( [ "foo", "bar", "baz", "qux" ] ), 
  [ "foo" ]
);

car(cons(X, Y)) == X
cdr(cons(X, Y)) == Y

If X is a string, then car returns a string. if X is a cons, then car needs to return a cons. Likewise with cdr and Y.

cowboy: if s were a list, then you are almost correct. list( cons( "foo", "bar" ), cons( "alpha", "beta" ) ) would return [ [ "foo", "bar" ], [ ["alpha", "beta" ], undefined ] ]. A list is made of conses, but not all conses make proper lists.

var l = list( cons( "foo", "bar" ), cons( "alpha", "beta" ) );
console.log( "list", l, [ [ "foo", "bar" ], [ ["alpha", "beta" ], undefined ] ] );
console.log( "car", car( l ), [ "foo", "bar" ] );
console.log( "cdr", cdr( l ), [ [ "alpha", "beta" ] ] );
console.log( "caar", caar( l ), "foo" );
console.log( "cadr", cadr( l ), [ "alpha", "beta" ] );
console.log( "caar", cdar( l ), "bar" );
console.log( "cadr", cddr( l ), undefined );

Of course, you need a working cons before you can even think of writing list. :)

Gotcha. I was unaware of the distinction between conses and lists, I did some research, and it makes sense now. Thanks! So, what about this, note how I changed the "cdr" test. Is this correct?

var l = list( cons( "foo", "bar" ), cons( "alpha", "beta" ) );
console.log( "list", l, [ [ "foo", "bar" ], [ ["alpha", "beta" ], undefined ] ] );
console.log( "car", car( l ), [ "foo", "bar" ] );
console.log( "cdr", cdr( l ), [ [ "alpha", "beta" ], undefined ] );
console.log( "caar", caar( l ), "foo" );
console.log( "cadr", cadr( l ), [ "alpha", "beta" ] );
console.log( "cdar", cdar( l ), "bar" );
console.log( "cddr", cddr( l ), undefined );

cowboy: yes, good catch. Lisp of course hides the nil at the end of every list, but it is still there