Missing return in Swift - swift

Question

Here is my code.
import UIKit
var str = "Hello, playground"
//There are two sorted arrays nums1 and nums2 of size m and n respectively.
//Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
//Example 1:
//nums1 = [1, 3]
//nums2 = [2]
//
//The median is 2.0
//Example 2:
//nums1 = [1, 2]
//nums2 = [3, 4]
//
//The median is (2 + 3)/2 = 2.5
var num1 = [1,2,2,5]
var num2 = [2,3,9,9]
class Solution {
func findMedianSortedArrays(_ nums1: [Int], _ nums2: [Int]) -> Double {
var A = nums1
var B = nums2
var m = nums1.count
var n = nums2.count
var max_of_left : Int = 0
var min_of_right = 0
if n < m {
var temp : [Int]
var tempt : Int
temp = nums1
tempt = m
A = nums2
B = temp
m = n
n = tempt
}
if n == 0{
fatalError("Arrays must be fulfilled")
}
var imin = 0
var imax = m
let half_len = Int((m+n+1)/2)
while imin <= imax {
let i = Int((imin + imax) / 2)
let j = half_len - i
if i > 0 && A[i-1] > B[j]{
imax = i - 1
}
else if i < m && A[i] < B[j-1]{
imin = i + 1
}
else
{
if i == 0{
max_of_left = B[j-1]
}
else if j == 0{
max_of_left = A[i-1]
}
else
{
max_of_left = max(A[i-1], B[j-1])
}
if m+n % 2 == 1{
return Double(max_of_left)
}
if i==m{
min_of_right = B[j]
}
else if j == n{
min_of_right = A[i]
}
else{
min_of_right = min(A[i], B[j])
//editor indicates error here
}
return Double((Double(max_of_left+min_of_right) / 2.0))
}
}
}
}
var a = Solution()
print(a.findMedianSortedArrays(num1, num2))
error: day4_Median_of_Two_Sorted_Arrays.playground:86:13: error: missing return in a function expected to return 'Double'
Since I put my return out of if statement, I think it will be okay because it will stop while looping when it meets return.
But editor says it's not.
I want to know why. Please explain me why.

Answer 1

Every code path through your findMedianSortedArrays() must return a Double.
So you need a return of a Double placed outside of your while loop. Even if you had every code path within the while loop have a return double, if imin > imax you wouldn't even enter the while loop, and so would need a return of a double outside it.

Answer 2

I fixed it by putting another return out of while loop.
//: Playground - noun: a place where people can play
import UIKit
var str = "Hello, playground"
//There are two sorted arrays nums1 and nums2 of size m and n respectively.
//Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
//Example 1:
//nums1 = [1, 3]
//nums2 = [2]
//
//The median is 2.0
//Example 2:
//nums1 = [1, 2]
//nums2 = [3, 4]
//
//The median is (2 + 3)/2 = 2.5
var num1 = [1,2,2,5]
var num2 = [2,3,9,9]
class Solution {
func findMedianSortedArrays(_ nums1: [Int], _ nums2: [Int]) -> Double {
var A = nums1
var B = nums2
var m = nums1.count
var n = nums2.count
var max_of_left : Int = 0
var min_of_right = 0
if n < m {
var temp : [Int]
var tempt : Int
temp = nums1
tempt = m
A = nums2
B = temp
m = n
n = tempt
}
if n == 0{
fatalError("Arrays must be fulfilled")
}
var imin = 0
var imax = m
let half_len = Int((m+n+1)/2)
while imin <= imax {
let i = Int((imin + imax) / 2)
let j = half_len - i
if i > 0 && A[i-1] > B[j]{
imax = i - 1
}
else if i < m && A[i] < B[j-1]{
imin = i + 1
}
else
{
if i == 0{
max_of_left = B[j-1]
}
else if j == 0{
max_of_left = A[i-1]
}
else
{
max_of_left = max(A[i-1], B[j-1])
}
if m+n % 2 == 1{
return Double(max_of_left)
}
if i==m{
min_of_right = B[j]
}
else if j == n{
min_of_right = A[i]
}
else{
min_of_right = min(A[i], B[j])
}
return Double((Double(max_of_left+min_of_right) / 2.0))
}
}
return Double((Double(max_of_left+min_of_right) / 2.0))
}
}
var a = Solution()
print(a.findMedianSortedArrays(num1, num2))