在递归中不返回任何内容的 return 语句

Question

I found this code about recursion online

function countDownRecursive(n) {
  if (n <= 0) {
    console.log('Hooray')
    return
  }

  console.log(n)
  countDownRecursive(n - 1)
}

I am really confused about this code, why does it console.log("Hooray") and then return nothing? Can you explain it to me? Thank you so much.

Answer 1

you returned a null value, the function output type is void. try this

if (n <= 0) {
    console.log('Hooray')
    return n
  }

Answer 2

return in this context means you don't want to continue running the function (similar to break in iterations).

The above recursive function's logic can be converted to this below while logic.

 let n = 3; //iterate until found the while break while (true) { //the condition to stop if (n <= 0) { console.log('Hooray'); break; //stop `while` } console.log(n) n = n - 1; }

Answer 3

why does it console.log("Hooray")

Because the function is recursive and when you start with let's say n=1 the function will not print "Hooray" immediately, because the condition:

if (n <= 0)

does not apply ie is false .

By the time we reach the recursion:

countDownRecursive(n - 1)

We call the function again with n=0 due to n - 1 , the if-statement will evaluate to true and therefore print "Hooray".

and then return nothing

It does not actually return "nothing", even though the return type is void , it returns undefined , which is the default behavior for return you could also write return undefined instead.

When you use return , it will basically terminate or return from the current function. It will jump back into the scope where you did call the function initially.

Hope that clears it up for you.

Answer 4

Let's do it with an example. This is the hierarchy of the calls when n = 5

countDownRecursive(5) // "5"
  countDownRecursive(4) // "4"
    countDownRecursive(3) // "3"
      countDownRecursive(2) // "2"
        countDownRecursive(1) // "1"
          countDownRecursive(0) // "Hooray" because n == 0, we execute the return statement
          end of countDownRecursive(0) because of return
        end of countDownRecursive(1) because reaching the end
      end of countDownRecursive(2) because reaching the end
    end of countDownRecursive(3) because reaching the end
  end of countDownRecursive(4) because reaching the end
end of countDownRecursive(5) because reaching the end

The returns statement tells the program to stop calling itself