[3,4,6,7][1,2,3] = 7? Javascript Tricky Interview question explained by Frontend Master  

array[index] // index (means one) but not indices/indexes let r = [9, 8, 7, 6][1, 2, 3, 0]; // [array][index] index = 0 1 2 3 // starts with 0 and the increment is 1 console.log(r); When we use the array [1, 2, 3, 0] as an index to access the elements in another array [9, 8, 7, 6], JavaScript does not throw a syntax error but it rather evaluates [1, 2, 3, 0] as an array expression. JS only considers the last element a ‘0’ as the index to access the element in the array [9, 8, 7, 6], and the other elements 1, 2, and 3 are ignored in this context. Output: 9 // Index considered by JS is only the last element, the right extreme One, in this case 0. let r = [9, 8, 7, 6][3] // Output 6 as element at index 3 is 6 index = 0 1 2 3 let r = [9, 8, 7, 6][1, 3] // Output 6 as element at index 3 is 6 let r = [9, 8, 7, 6][1, 0, 3] // Output 6 as element at index 3 is 6 let r = [9, 8, 7, 6][1, 0, 2, 3] // Output 6 as element at index 3 is 6 index = 0 1 2 3 If we need more than one element, use slice(start_Index_include, end_Index_exclude) let r = [9, 8, 7, 6].slice(1, 3); console.log(r) Output: (2) [8, 7] //------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- function a() {return "a"} function b() {return "b"} function c() {return "c"} console.log(a(), b(), c()); console.log(c(), a(), b()); console.log(b(), c(), a()); Output: a b c c a b b c a

Brilliant...

Keep going sir . I like the way you teach and do some xss attack in real time

could you please answer what will be output for this let b = [14,15,8,115][1,2,3][1]; console.log(b); as per your statement the value of b should be 8 but it is undefind why??

what happen with tthe 3 then ?

those who are facing pop-up on Answer. Try inspect element and make VISIBILITY:HIDDEN