Execute function in command pattern JavaScript -

i in mist of reading addy osmani's javascript design patterns can found here: http://addyosmani.com/resources/essentialjsdesignpatterns/book/.

i finding quite enjoyable , helpful. have question 1 of patterns in book, namely command pattern.

in book, addy explains command pattern helpful decouple objects , method calls little better.

here version of example:

var person = {     sayname: function (name) {         return "my name " + name;     },     sayage: function (age) {         return "my age " + age;     },     saygender: function (gender) {         return "my gender " + gender;     } } person.execute = function (name) {     return person[name] && person[name].apply(person, [].slice.call(arguments, 1)); } console.log(person.execute("sayname", "sethen")); 

the magic done in execute method. can see, pass method name , arguments , method takes care of rest.

my confusion stems execute method returning. when @ it, looks short circuit && thought returned boolean due javascript conversion.

however, if try code works should, logging my name sethen.

furthermore, i've found using return person[name].apply(person, [].slice.call(arguments, 1); produces same results , in mind easier read.

so, question how return works in original execute method:

return person[name] && person[name].apply(person, [].slice.call(arguments, 1)); 

and how && operator works in instance make work??

the && operator not return boolean. returns value of last subexpression evaluates. if requested method name exists, expression return result of .apply() call.

so:

return person[name] && person[name].apply(person, [].slice.call(arguments, 1)); 

means:

if (!person[name])   return person[name];  return person[name].apply(person, [].slice.call(arguments, 1));