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Start your free trialArikaturika Tumojenko
8,897 PointsDid I get this correctly (what's the role of 'songs' inside the function)?
I had this issue before, when creating the first example of displaying the playlist using one dimensional arrays. I have this piece of code
function printSongs( songs ) {
var listHTML = '<ol>';
for ( var i = 0; i < songs.length; i += 1) {
listHTML += "<li>" + songs[i][0] + "by" + songs[i][1] + "</li>";
}
listHTML += '</ol>';
print(listHTML);
}
printSongs(playList);
Is this the same as writing the code like this?
function printSongs( playList ) {
var listHTML = '<ol>';
for ( var i = 0; i < playList.length; i += 1) {
listHTML += "<li>" + playList[i][0] + "by" + playList[i][1] + "</li>";
}
listHTML += '</ol>';
print(listHTML);
}
printSongs(playList);
If so, it means that the placeholder "songs" was used just for creating the function and it has no other role?
Did Dave changed the value "list" with the value "song" to be more adequate as a name or it has other implications? Thank you!
1 Answer
Tom Byers
13,005 PointsHi Arikaturika,
I'm not sure of the specific context here but I hope I can help.
The use of both songs
and playList
in the function are simply arguments that are passed into the function. They act as placeholders, as you say. They could be given pretty well any name you like. For example:
// create a function called printSongs:
function printSongs(anArray) {
var listHTML = '<ol>';
for (var i = 0; i < anArray.length; i += 1) {
listHTML += "<li>" + anArray[i][0] + "by" + anArray[i][1] + "</li>";
}
listHTML += '</ol>';
print(listHTML);
}
// call the function, passing in your actual array (e.g. songs, playList etc).:
printSongs(songsList);
I think you've got it. It's just a placeholder, ready for when you call the function with the actual variable name.