JavaScript array_unshift
Pushes elements onto the beginning of the array
1 2 3 4 56 7 8 9 1011 12 13 14 1516 17 18 19 | function array_unshift (array) { // Pushes elements onto the beginning of the array // // version: 1008.1718 // discuss at: http://phpjs.org/functions/array_unshift // + original by: Kevin van Zonneveld (http://kevin.vanzonneveld.net) // + improved by: Martijn Wieringa // + improved by: jmweb // % note 1: Currently does not handle objects // * example 1: array_unshift(['van', 'Zonneveld'], 'Kevin'); // * returns 1: 3 var i = arguments.length; while(--i !== 0){ arguments[0].unshift(arguments[i]); } return arguments[0].length; } |
Examples
Running
1 | array_unshift(['van', 'Zonneveld'], 'Kevin'); |
Should return
1 | 3 |
Dependencies
No dependencies, you can use this function standalone.
Open syntax issues
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Authors
Thanks to the following developers, you get to have array_unshift goodness in JavaScript.
jmweb
May 26th 
Note that the array_unshift() implementation given here prepends the array arguments in the wrong order.
As an example:
var names=['van', 'Zonneveld']; array_unshift(names,'Kevin','Tim') // returns 4 //names is now ['Tim','Kevin','van', 'Zonneveld']
The function should be altered to:
function array_unshift(/* assumes at least 1 argument passed - the array */){
var i=arguments.length;
while(--i!==0){
arguments[0].unshift(arguments[i]);
}
return arguments[0].length;
}
As an example:
var names=['van', 'Zonneveld']; array_unshift(names,'Kevin','Tim') // returns 4 //names is now ['Kevin','Tim','van', 'Zonneveld']
Kevin van Zonneveld
Jun 19th