I am a little confused with this piece of code. I would really appreciate a step by step walk-through.
i = 0;
!(i % 3 && i % 5) && arr.push(i), i++
Asked
Active
Viewed 28 times
-3
-
To clarify - what don't you understand about it? – Oliver Charlesworth Feb 19 '18 at 22:02
-
not ( even % 3 && even % 3 ) and (so do) arr.push(i), i= i+1 – Abdelrahman M. Allam Feb 19 '18 at 22:03
-
`i = 0; if (i % 3 == 0 || i % 5 == 0) { arr.push(i); } i++;` – GSerg Feb 19 '18 at 22:04
-
1&& operator is evaluate the sides of conditions true && true , false&& true , so if !(i % 3 && i % 5) it will go to evaluate && arr.push(i) and , is concatenation for declaration next statement for separation – Abdelrahman M. Allam Feb 19 '18 at 22:07
-
Possible duplicate of [Javascript AND operator within assignment](https://stackoverflow.com/questions/3163407/javascript-and-operator-within-assignment) – CodeLover Feb 19 '18 at 22:10
1 Answers
0
A step by step walkthrough:
i = 0; // assignment of 0 to the global variable i//
!// the opposite of the following expression, note does not apply to
//expression not in parentheses below //
(i % 3 && i % 5) // returns
//true if i does
// equal zero for both i % 3 and i % 5 // in javascript 0 is a falsy
//value
&& arr.push(i) // returns true if the length property of arr is not
// 0, i++
Matthew Lagerwey
- 150
- 6