Trevor wishes to duplicate an array, transforming it by multiplying its elements four times.

Existing Solution

var atest =   ['a','b','c'];
var btest =   [];
btest     =   atest.concat(atest).concat(atest).concat(atest);
console.log(btest.length);  // btest.length == 12
console.log(btest);         // ["a", "b", "c", "a", "b", "c", "a", "b", "c", "a", "b", "c"] 


Trevor does not like this because it is ugly. Can you help Trevor find his happy place again? It's bugging Trevor.

  • 13
    \$\begingroup\$ Trevor should come here and post the question himself so he can provide some context as to why he wants to multiply the array's elements 4 times. \$\endgroup\$
    – Pimgd
    Oct 28, 2014 at 13:55
  • \$\begingroup\$ Will it always be 4, might a loop not make sense? \$\endgroup\$
    – konijn
    Oct 28, 2014 at 17:40
  • \$\begingroup\$ //why he wants to multiply the array's elements 4 times// Why not? \$\endgroup\$
    – dreftymac
    Oct 29, 2014 at 14:57
  • \$\begingroup\$ //Will it always be 4, might a loop not make sense?// A loop is fine, but something more like the way this can be done in ruby is preferred: print (['a','b','c'] * 4).inspect that gives the desired outcome \$\endgroup\$
    – dreftymac
    Oct 29, 2014 at 15:10
  • 1
    \$\begingroup\$ I am afraid that Trevor would need to write a helper function for that ;) \$\endgroup\$
    – konijn
    Oct 30, 2014 at 13:26

1 Answer 1


Here's one that uses call and manually concat a

var a = ['a','b','c'], b = Array.prototype.concat.call(a,a,a,a);

If you wanted an easy way to variably add a, just change i:

for(var a=["a","b","c"],b=[],i=4;i--;) b.push.apply(b,a);

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