--- id: 587d7b89367417b2b2512b48 title: Use the Spread Operator to Evaluate Arrays In-Place challengeType: 1 forumTopicId: 301222 dashedName: use-the-spread-operator-to-evaluate-arrays-in-place --- # --description-- ES6 introduces the spread operator, which allows us to expand arrays and other expressions in places where multiple parameters or elements are expected. The ES5 code below uses `apply()` to compute the maximum value in an array: ```js var arr = [6, 89, 3, 45]; var maximus = Math.max.apply(null, arr); ``` `maximus` would have a value of `89`. We had to use `Math.max.apply(null, arr)` because `Math.max(arr)` returns `NaN`. `Math.max()` expects comma-separated arguments, but not an array. The spread operator makes this syntax much better to read and maintain. ```js const arr = [6, 89, 3, 45]; const maximus = Math.max(...arr); ``` `maximus` would have a value of `89`. `...arr` returns an unpacked array. In other words, it *spreads* the array. However, the spread operator only works in-place, like in an argument to a function or in an array literal. The following code will not work: ```js const spreaded = ...arr; ``` # --instructions-- Copy all contents of `arr1` into another array `arr2` using the spread operator. # --hints-- `arr2` should be correct copy of `arr1`. ```js assert(arr2.every((v, i) => v === arr1[i]) && arr2.length); ``` `...` spread operator should be used to duplicate `arr1`. ```js assert(code.match(/Array\(\s*\.\.\.arr1\s*\)|\[\s*\.\.\.arr1\s*\]/)); ``` `arr2` should remain unchanged when `arr1` is changed. ```js assert((arr1, arr2) => { arr1.push('JUN'); return arr2.length < arr1.length; }); ``` # --seed-- ## --seed-contents-- ```js const arr1 = ['JAN', 'FEB', 'MAR', 'APR', 'MAY']; let arr2; arr2 = []; // Change this line console.log(arr2); ``` # --solutions-- ```js const arr1 = ['JAN', 'FEB', 'MAR', 'APR', 'MAY']; let arr2; arr2 = [...arr1]; ```