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Home/ Questions/Q 7759965
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Editorial Team
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Editorial Team
Asked: June 1, 20262026-06-01T13:45:34+00:00 2026-06-01T13:45:34+00:00

Hi I have been trying to learn Javascript using codeacademy.com and I have reached

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Hi I have been trying to learn Javascript using codeacademy.com and I have reached an exercise that doesn’t seem to make any sense when why the exercise I have written.This is my code:

 (function(){
    var bob = {
        firstName: "Bob",
        lastName: "Jones",

        phoneNumber: "(650) 777 - 7777",
        email: "bob.jones@example.com"
    };

    var mary = {
        firstName: "Mary",
        lastName: "Johnson",

        phoneNumber: "(650) 888 - 8888",
        email: "mary.johnson@example.com"
    };

    var contacts = [bob, mary];

    var printPerson = function(person){
        console.log(person.firstName + " " + person.lastName);
    }

    var list = function(){
        var i = contacts.length;
        for(var j= 0; j < i ; j++){
            printPerson(contacts[i]);
        }
    };

        list();
})();

The problem is in the list function when I try to call the printPerson() function I get that person is undefined but if I write instead of the list() function this:

    printPerson(contacts[0]);
    printPerson(contacts[1]);

Everything works.What am I doing wrong in the list() function that it doesn’t work?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-01T13:45:35+00:00Added an answer on June 1, 2026 at 1:45 pm
    var list = function(){
        var i = contacts.length;
        for(var j= 0; j < i ; j++){
            printPerson(contacts[i]);
        }
    };
    

    i here is a constant. If you replace it:

    var list = function(){
        var i = contacts.length;
        for(var j= 0; j < contacts.length ; j++){
            printPerson(contacts[contacts.length]);
        }
    };
    

    For all arrays arr, arr[arr.length] will always be undefined. You probably want contacts[j].

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