你如何在 swift 中使用 reduce(into:)

Question

I am reading iOS 13 Programming Fundamentals with Swift, got to the part about reduce() and I think I understand it more or less, but then there is reduce(into:) and this piece of code:

let nums = [1,2,3,4,5]
let result = nums.reduce(into: [[],[]]) { temp, i in 
    temp[i%2].append(i)
}
// result is now [[2,4],[1,3,5]]

So this code takes an array of Int and splits it into 2 arrays, even and odd. The problem is that I have no idea what's happening inside the brackets {}.

In the case of reduce, the first parameter is the first one of the iteration and then the closure is supposed to process all the items one after the other, similar to map() but more powerful (here one loop is enough to get the two arrays but with map() I would need 2 loops, according to the book).

I cannot understand the syntax here anyway, especially what does "temp" stand for and that use of "in". And how is "append()" appending the value to the proper array??

Answer 1

Inside the closure, "temp" is the result format which is [[][]] and "i" is each number. As you said it processes all numbers in a loop. When % is used it returns the division remainder, so for the odd numbers like "1,3,5", it returns "1" and for the even numbers "0", which means that "temp" appends these values to the array in these respective indexes.

So if we debug and replace the variables for constants the results would be:

temp[1].append(1) //1%2 = 1/2 left 1 [[][1]]
temp[0].append(2) //2%2 = 2/2 left 0 [[2][1]]
temp[1].append(3) //3%2 = 3/2 = 1 left 1 [[2][1,3]]
temp[0].append(4) //4%2 = 4/2 left 0 [[2,4][1,3]]
temp[1].append(5) //5%2 = 5/2 = 2 left 1 [[2,4][1,3,5]]

According to the documentation the closure is called sequentially with a mutable accumulating value initialized that when exhausted, is returned to the caller.