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Start your free trialNatalia Karolinskaya
9,367 PointsI don't understand two things in this exaple
- listHTML += '<li>' + list[i] + '</li>';
Initial value of i is 0, so ho come the list starts with number 1?
- I don't see where in the function printList we instruct it to print the playList? I mean we say print playList in the end, but that is note the function code anymore.
Thanks for your help!
2 Answers
Chyno Deluxe
16,936 PointsAn arrays index begins at 0 so that is why the inital value of i = 0.
var playList = [
'I Did It My Way', // playList[0]
'Respect', // playList[1]
'Imagine', // playList[2]
'Born to Run', // playList[3]
'Louie Louie', // playList[4]
'Maybellene' // playList[5]
];
So the printList function cycles through the playlist array and prints them out.
function printList( list ) { // list = playList[]
var listHTML = '<ol>';
//This for loop prints out each item in the playList array
for (var i = 0; i < list.length; i += 1) {
listHTML += '<li>' + list[i] + '</li>';
}
listHTML += '</ol>';
print(listHTML);
}
printList(playList); // playList is called as an argument in the printList function
john larson
16,594 Pointsalright, I have the same questions as noted above...and I will raise you this one. WHY does list = playList. That is not mentioned thus far that I can see. If it is an understood shortcut of some sort, I would have been less confused if that were mentioned in the lesson.
Chyno Deluxe
16,936 Pointslist = playList because you call the printList() function with playList as the argument.
printList(playList);
The parameter list could be called anything and It would still equal to playList[]. View Below
function printList( k ) { // k = playList[]
var listHTML = '<ol>';
//This for loop prints out each item in the playList array
for (var i = 0; i < k.length; i += 1) {
listHTML += '<li>' + k[i] + '</li>';
}
listHTML += '</ol>';
print(listHTML);
}
printList(playList); // playList is called as an argument in the printList function
I hope this helps. Feel free to ask me another other questions you may have.