How to get the Power of some Integer in Swift language

Question

I m learning swift recently  but I have a basic problem that can t find an answer  I want to get something like   var a Int   3 var b Int   3  println  pow a b       27   but the pow function can work with double number only  it doesn t work with integer  and I can t even cast the int to double by something like Double a  or a double       Why it doesn t supply the power of integer  it will definitely return an integer without ambiguity   and Why I can t cast a integer to a double  it just change 3 to 3 0  or 3 00000    whatever   if I got two integer and I want to do the power operation  how can I do it smoothly   Thanks

User · Answer

To calculate power 2  n   simply use   let result   2  lt  lt   n-1

User · Answer

It turns out you can also use pow    For example  you can use the following to express 10 to the 9th   pow 10  9    Along with pow  powf   returns a float instead of a double  I have only tested this on Swift 4 and macOS 10 13

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An Int-based pow function that computes the value directly via bit shift for base 2 in Swift 5   func pow base  Int  power  UInt  - gt  Int       if power    0   return 1          for base 2  use a bit shift to compute the value directly     if base    2   return 2  lt  lt  Int power - 1           otherwise multiply base repeatedly to compute the value     return repeatElement base  count  Int power   reduce 1          Make sure the result is within the range of Int - this does not check for the out of bounds case

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If you really want an  Int only  implementation and don t want to coerce to from Double  you ll need to implement it   Here is a trivial implementation  there are faster algorithms but this will work  func pow    base Int    power UInt  - gt  Int     var answer   Int   1   for   in 0   lt power   answer    base     return answer     gt  pow  2  4   R3  Int   16  gt  pow  2  8   R4  Int   256  gt  pow  3 3   R5  Int   27  In a real implementation you d probably want some error checking

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In Swift 5   extension Int      func expo   power  Int  - gt  Int           var result   1         var powerNum   power         var tempExpo   self         while  powerNum    0           if  powerNum 2    1               result    tempExpo                   powerNum    2         tempExpo    tempExpo                   return result            Use like this   2 expo 5     pow 2  5    Thanks to  Paul Buis s answer

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If you re disinclined towards operator overloading  although the    solution is probably clear to someone reading your code  you can do a quick implementation   let pwrInt  Int Int - gt Int     a b in return Int pow Double a  Double b      pwrInt 3 4     81

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I like this better  func    left NSNumber  right  NSNumber  - gt  NSNumber       return pow left doubleValue right doubleValue    var a NSNumber   3 var b NSNumber   3  println  a b      27

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Or just     var a Int   3 var b Int   3 println pow Double a  Double b

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Swift 4 x version  precedencegroup ExponentiationPrecedence     associativity  right   higherThan  MultiplicationPrecedence    infix operator     ExponentiationPrecedence public func     radix  Float  power  Float  - gt  Float     return pow  radix    power      public func     radix  Double  power  Double  - gt  Double     return pow  radix    power      public func     radix  Int  power  Int  - gt  Int     return NSDecimalNumber decimal  pow Decimal radix   power   intValue

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Trying to combine the overloading  I tried to use generics but couldn t make it work   I finally figured to use NSNumber instead of trying to overload or use generics   This simplifies to the following   typealias Dbl   Double    Shorter form infix operator     associativity left precedence 160  func     lhs  NSNumber  rhs  NSNumber  - gt  Dbl  return pow Dbl lhs   Dbl rhs      The following code is the same function as above but implements error checking to see if the parameters can be converted to Doubles successfully   func     lhs  NSNumber  rhs  NSNumber  - gt  Dbl          Added  probably unnecessary  check that the numbers converted to Doubles     if  Dbl lhs     Dbl NaN     Dbl NaN  amp  amp   Dbl rhs     Dbl NaN     Dbl NaN           return pow Dbl lhs   Dbl rhs         else           return Double NaN

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Combining the answers into an overloaded set of functions  and using      instead of      as some other languages use - clearer to me       http   stackoverflow com questions 24196689 how-to-get-the-power-of-some-integer-in-swift-language    Put this at file level anywhere in your project infix operator      associativity left precedence 160   func     radix  Double  power  Double  - gt  Double   return pow radix  power    func     radix  Int     power  Int     - gt  Double   return pow Double radix   Double power     func     radix  Float   power  Float   - gt  Double   return pow Double radix   Double power       When using Float  you may lose precision   If using numeric literals and a mix of integers and non-integers  you will end up with Double by default   I personally like the ability to use a mathematical expression instead of a function like pow a  b  for stylistic readability reasons  but that s just me   Any operators that would cause pow   to throw an error will also cause these functions to throw an error  so the burden of error checking still lies with the code using the power function anyway   KISS  IMHO   Using the native pow   function allows to eg take square roots  2    0 5  or inverse  2    -3   1 8    Because of the possibility to use inverse or fractional exponents  I wrote all my code to return the default Double type of the pow   function  which should return the most precision  if I remember the documentation correctly    If needed  this can be type-casted down to Int or Float or whatever  possibly with the loss of precision   2    -3    0 125 2    0 5   1 4142135623731 2    3     8

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The other answers are great but if preferred  you can also do it with an Int extension so long as the exponent is positive  extension Int          func pow toPower  Int  - gt  Int           guard toPower  gt  0 else   return 0           return Array repeating  self  count  toPower  reduce 1              2 pow toPower  8     returns 256

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Other than that your variable declarations have syntax errors  this works exactly how you expected it to  All you have to do is cast a and b to Double and pass the values to pow  Then  if you re working with 2 Ints and you want an Int back on the other side of the operation  just cast back to Int   import Darwin   let a  Int   3 let b  Int   3  let x  Int   Int pow Double a  Double b

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If you like  you could declare an infix operator to do it      Put this at file level anywhere in your project infix operator      associativity left precedence 160   func     radix  Int  power  Int  - gt  Int       return Int pow Double radix   Double power                 Then you can do this    let i   2    3        or println  2       2    3       Prints 2     8   I used two carets so you can still use the XOR operator   Update for Swift 3  In Swift 3 the  magic number  precedence is replaced by precedencegroups   precedencegroup PowerPrecedence   higherThan  MultiplicationPrecedence   infix operator      PowerPrecedence func     radix  Int  power  Int  - gt  Int       return Int pow Double radix   Double power                 Then you can do this    let i2   2    3        or print  2       2    3       Prints 2     8

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Array repeating  a  count  b  reduce 1

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Swift 5  I was surprised  but I didn t find a proper correct solution here   This is mine   enum CustomMath lt T  BinaryInteger gt         static func pow   base  T    power  T  - gt  T           var tempBase   base         var tempPower   power         var result  T   1          while  power    0                if  power   2    1                    result    base                           tempPower   tempPower  gt  gt  1             tempBase    tempBase                   return result           Example   CustomMath pow 1 1

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Sometimes  casting an Int to a Double is not a viable solution  At some magnitudes there is a loss of precision in this conversion  For example  the following code does not return what you might intuitively expect   Double Int max - 1   lt  Double Int max     false    If you need precision at high magnitudes and don t need to worry about negative exponents     which can t be generally solved with integers anyway     then this implementation of the tail-recursive exponentiation-by-squaring algorithm is your best bet  According to this SO answer  this is  the standard method for doing modular exponentiation for huge numbers in asymmetric cryptography       using Swift 5 0 func pow lt T  BinaryInteger gt    base  T    power  T  - gt  T       func expBySq   y  T    x  T    n  T  - gt  T           precondition n  gt   0          if n    0               return y           else if n    1               return y   x           else if n isMultiple of  2                return expBySq y  x   x  n   2            else      n is odd             return expBySq y   x  x   x   n - 1    2                       return expBySq 1  base  power       Note  in this example I ve used a generic T  BinaryInteger  This is so you can use Int or UInt or any other integer-like type

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mklbtz is correct about exponentiation by squaring being the standard algorithm for computing integer powers  but the tail-recursive implementation of the algorithm seems a bit confusing  See http   www programminglogic com fast-exponentiation-algorithms  for a non-recursive implementation of exponentiation by squaring in C  I ve attempted to translate it to Swift here   func expo   base  Int    power  Int  - gt  Int       var result   1      while  power    0           if  power 2    1               result    base                   power    2         base    base           return result     Of course  this could be fancied up by creating an overloaded operator to call it and it could be re-written to make it more generic so it worked on anything that implemented the IntegerType protocol  To make it generic  I d probably start with something like      func expo lt T IntegerType gt    base  T    power  T  - gt  T       var result   T   1   But  that is probably getting carried away

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little detail more      infix operator      associativity left precedence 160      func     radix  Int  power  Int  - gt  Int          return Int pow CGFloat radix   CGFloat power           swift - Binary Expressions

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func calc  base Int  number Int  - gt  Int       var answer   Int   base     for   in 2   number  answer    base        return answer     Example   calc  2 2

[integer] How to get the Power of some Integer in Swift language?

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