Find digit sum of a number in Swift - swift

Question

Does anyone know how to get the sum of all the digits in a number in Swift?
For example using the number 845 would result in 17

Answer 1

update: Swift 5 or later We can use then new Character property wholeNumberValue:
let string = "845"
let sum = string.compactMap{$0.wholeNumberValue}.reduce(0, +)
print(sum) // 17
let integer = 845
let sumInt = String(integer).compactMap{$0.wholeNumberValue}.reduce(0, +)
print(sumInt) // 17

Answer 2

Here is a solution that uses simple integer arithmetic only:
func digitSum(var n : Int) -> Int {
var sum = 0
while n > 0 {
sum += n % 10 // Add least significant digit ...
n /= 10 // ... and remove it from the number.
}
return sum
}
println(digitSum(845)) // 17
Update for Swift 3/4:
func digitSum(_ n : Int) -> Int {
var n = n
var sum = 0
while n > 0 {
sum += n % 10 // Add least significant digit ...
n /= 10 // ... and remove it from the number.
}
return sum
}
print(digitSum(845)) // 17
Another implementation, just for fun:
func digitSum(_ n : Int) -> Int {
return sequence(state: n) { (n: inout Int) -> Int? in
defer { n /= 10 }
return n > 0 ? n % 10 : nil
}.reduce(0, +)
}

Answer 3

The recursive solution in Swift 3!
func digitSum(of number: Int) -> Int {
if(number < 10) {
return number
} else {
return (number % 10) + digitSum(of: (number/10))
}
}

Answer 4

For the sake of completeness, and for those who would like to see or understand a math-based approach, here's a real-number function based technique ported to Swift.
This is not the most efficient way to tally the digits of an integer in Swift. I don't recommend using it. I would personally use #LeoLDbus map/reduce answer to the question, because it is so cool and illustrates a powerful set of Swift features yet short, or #MartinR integer mod/divide answer, due to its utter simplicity and relative speed of integer arithmetic .
Cocoa and UIKit have the requisite math methods, so you'll probably need to import one of those packages.
func sumDigits(var i : Int) -> Int {
var sum = 0
var nDigits = floor(log10(Double(i))) + 1
for var r = nDigits; r > 0; r-- {
var p = pow(10, r - 1)
var d = floor(Double(i) / p)
sum += Int(d)
i -= Int(d * p)
}
return sum
}

Answer 5

for swift4, try below function:
func sumDigits(num: Int) -> Int {
return String(num).compactMap { Int(String($0)) }.reduce(0, +)
}

Answer 6

Split it into two pieces:
digits
public extension UnsignedInteger {
/// The digits that make up this number.
/// - Parameter radix: The base the result will use.
func digits(radix: Self = 10) -> [Self] {
sequence(state: self) { quotient in
guard quotient > 0
else { return nil }
let division = quotient.quotientAndRemainder(dividingBy: radix)
quotient = division.quotient
return division.remainder
}
.reversed()
}
}
XCTAssertEqual(
(867_5309 as UInt).digits(),
[8,6,7, 5,3,0,9]
)
XCTAssertEqual(
(0x00_F0 as UInt).digits(radix: 0b10),
[1,1,1,1, 0,0,0,0]
)
XCTAssertEqual(
(0xA0_B1_C2_D3_E4_F5 as UInt).digits(radix: 0x10),
[10,0, 11,1, 12,2, 13,3, 14,4, 15,5]
)
XCTAssertEqual(
(0o00707 as UInt).digits(radix: 0o10),
[0b111, 0, 0b111]
)
sum
public extension Sequence where Element: AdditiveArithmetic {
var sum: Element? { reduce(+) }
}
public extension Sequence {
/// The first element of the sequence.
/// - Note: `nil` if the sequence is empty.
var first: Element? {
var iterator = makeIterator()
return iterator.next()
}
/// - Returns: `nil` If the sequence has no elements, instead of an "initial result".
func reduce(
_ getNextPartialResult: (Element, Element) throws -> Element
) rethrows -> Element? {
guard let first = first
else { return nil }
return try dropFirst().reduce(first, getNextPartialResult)
}
}
XCTAssertEqual([1, 2, 3].sum, 6)
XCTAssertEqual([0.5, 1, 1.5].sum, 3)
XCTAssertNil([CGFloat]().sum)