ForEach loop not printing the 4 element of the array - JavaScript

Question

Here's my code -

 var a = 'abcde'.split(''); var b = 'cxefgh'.split(''); //abfgh a.forEach((e, index) => { console.log(e, index) if (b.indexOf(e) > -1) { b.splice(b.indexOf(e), 1); a.splice(index, 1); } }) //Here the output - //a 0 //b 1 //c 2 //e 3 

Why this loop isn't printing d. What's the silliest thing that I am missing here?

Answer 1

You are changing a while iterating on it, resulting in discrepancy.
You should consider working on clone of a (created with a.slice() ) and then changing the a and b , if matches are found.

 var a = 'abcde'.split(''); var b = 'cxefgh'.split(''); //abfgh a.slice().forEach((e , index)=> { console.log(e, index) if(b.indexOf(e) > -1) { b.splice(b.indexOf(e) , 1); // Find where it is in `b` and remove it. a.splice(a.indexOf(e) , 1); // Find where it is in `a` and remove it. } }) console.log(a); // ['a', 'b', 'd'] console.log(b); // ['x', 'f', 'g', 'h'] 

Answer 2

a.splice(index , 1); is executed when e is 'c' . index is 2 at that point, so a becomes ['a', 'b', 'd', 'e'] (the splice removes the 'c' ). Then the forEach moves on with index 3, which is (now) 'e' .