How exactly does the “let” keyword work in Swift?

Question

I\'ve read this simple explanation in the guide:

The value of a constant doesn’t need to be known at compile time, but you must assign it a value exa

Answer 1

It's best to think of let in terms of Static Single Assignment (SSA) -- every SSA variable is assigned to exactly once. In functional languages like lisp you don't (normally) use an assignment operator -- names are bound to a value exactly once. For example, the names y and z below are bound to a value exactly once (per invocation):

func pow(x: Float, n : Int) -> Float {
  if n == 0 {return 1}
  if n == 1 {return x}
  let y = pow(x, n/2)
  let z = y*y
  if n & 1 == 0 {
    return z
  }
  return z*x
}

This lends itself to more correct code since it enforces invariance and is side-effect free.

Here is how an imperative-style programmer might compute the first 6 powers of 5:

var powersOfFive = Int[]()
for n in [1, 2, 3, 4, 5, 6] {
    var n2 = n*n
    powersOfFive += n2*n2*n
}

Obviously n2 is is a loop invariant so we could use let instead:

var powersOfFive = Int[]()
for n in [1, 2, 3, 4, 5, 6] {
    let n2 = n*n
    powersOfFive += n2*n2*n
}

But a truly functional programmer would avoid all the side-effects and mutations:

let powersOfFive = [1, 2, 3, 4, 5, 6].map(
    {(num: Int) -> Int in
        let num2 = num*num
        return num2*num2*num})