Can anyone explain my "by mistake" solution ? - Algorithm Chunky Monkey

Can anyone explain my "by mistake" solution ? - Algorithm Chunky Monkey
0

#1

Hi everyone,

I was struggling to find a solution to this challenger when “by mistake” it worked; now I can’t understand the FOR LOOP logic here.

for ( var i = 0; arr.length; i+=size) // whats the meaning of this? Cause var i isn’t < or > or <= >= arr.length.

Best regards.
Paulo.

Entire Solution below .

`function chunkArrayInGroups(arr, size) {
var secondArray = [];

for ( var i = 0; arr.length; i+=size) {
secondArray.push(arr.slice(0, size));

//secondArray.push(arr.slice(size));

arr = arr.slice(size);

}

return secondArray;
}

chunkArrayInGroups([“a”, “b”, “c”, “d”], 2);


#2

Second parameter in for loops gets evaluated before each loop, if it evaluates to truthy, loop gets executed if it is falsy, loop stops.

In each loop you are changing the array arr by slicing it and every time arr.length gets smaller (first it’s 4, then 2 and then 0 which evaluates to false, so the loop stops).


#3

Thanks for your replied @jenovs ; I thought I was obliged to put

i < arr.lenght. // or i > arr.length

to be evaluated by > for loop . so just to clarify:

In the first time, > ( var i = 0; arr.length; i+=size)

var i = 0,
arr.lenght = 4 ----- “the four elements for the array” ;
and in each loop i ++ by size; ok…

so in this process it doesn’t matter the size of the var i; is this correct ? cause var i it’s not being evaluated.


#4

Correct!

You can check the syntax of a for loop and you’ll see that all the expressions are optional.

So this is a valid for loop:

var i = 0;
for ( ; ; ) {
  console.log(i)
  i++
  if (i > 5) break;
}

#5

Ok; I got it.

Thanks a lot @jenovs ; I appreciated your help.