Generate a random float between 0 and 1

Question

I'm trying to generate a random number that's between 0 and 1. I keep reading about arc4random(), but there isn't any information about getting a float from it. How do I do this?

Answer 1

// Seed (only once)
srand48(time(0));

double x = drand48();

// Swift version
// Seed (only once)
srand48(Int(Date().timeIntervalSince1970))

let x = drand48()

The drand48() and erand48() functions return non-negative, double-precision, floating-point values, uniformly distributed over the interval [0.0 , 1.0].

Answer 2

Swift 4.2+

Swift 4.2 adds native support for a random value in a Range:

let x = Float.random(in: 0.0...1.0)
let y = Double.random(in: 0.0...1.0)
let z = Float80.random(in: 0.0...1.0)

Doc:

• random(in range: ClosedRange<Float>)
• random(in range: Range<Float>)
• random(in range: ClosedRange<Double>)
• random(in range: Range<Double>)
• random(in range: ClosedRange<Float80>)
• random(in range: Range<Float80>)

Answer 3

How about this operation ((CGFloat)(rand()%100)/100) ?

Answer 4

float x = arc4random() % 11 * 0.1;

That produces a random float bewteen 0 and 1. More info here

Answer 5

For Swift 4.2+ see: https://stackoverflow.com/a/50733095/1033581

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Below are recommendations for correct uniformity and optimal precision for ObjC and Swift 4.1.

32 bits precision (Optimal for Float)

Uniform random value in [0, 1] (including 0.0 and 1.0), up to 32 bits precision:

Obj-C:

float val = (float)arc4random() / UINT32_MAX;

Swift:

let val = Float(arc4random()) / Float(UInt32.max)

It's optimal for:

• a Float (or Float32) which has a significand precision of 24 bits for its mantissa

48 bits precision (discouraged)

It's easy to achieve 48 bits precision with drand48 (which uses arc4random_buf under the hood). But note that drand48 has flaws because of the seed requirement and also for being suboptimal to randomize all 52 bits of Double mantissa.

Uniform random value in [0, 1], 48 bits precision:

Swift:

// seed (only needed once)
srand48(Int(Date.timeIntervalSinceReferenceDate))
// random Double value
let val = drand48()

64 bits precision (Optimal for Double and Float80)

Uniform random value in [0, 1] (including 0.0 and 1.0), up to 64 bits precision:

Swift, using two calls to arc4random:

let arc4random64 = UInt64(arc4random()) << 32 &+ UInt64(arc4random())
let val = Float80(arc4random64) / Float80(UInt64.max)

Swift, using one call to arc4random_buf:

var arc4random64: UInt64 = 0
arc4random_buf(&arc4random64, MemoryLayout.size(ofValue: arc4random64))
let val = Float80(arc4random64) / Float80(UInt64.max)

It's optimal for:

• a Double (or Float64) which has a significand precision of 52 bits for its mantissa
• a Float80 which has a significand precision of 64 bits for its mantissa

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Notes

Comparisons with other methods

Answers where the range is excluding one of the bounds (0 or 1) likely suffer from a uniformity bias and should be avoided.

• using arc4random(), best precision is 1 / 0xFFFFFFFF (UINT32_MAX)
• using arc4random_uniform(), best precision is 1 / 0xFFFFFFFE (UINT32_MAX-1)
• using rand() (secretly using arc4random), best precision is 1 / 0x7FFFFFFF (RAND_MAX)
• using random() (secretly using arc4random), best precision is 1 / 0x7FFFFFFF (RAND_MAX)

It's mathematically impossible to achieve better than 32 bits precision with a single call to arc4random, arc4random_uniform, rand or random. So our above 32 bits and 64 bits solutions should be the best we can achieve.

Answer 6

Swift 4.2

Swift 4.2 has included a native and fairly full-featured random number API in the standard library. (Swift Evolution proposal SE-0202)

let intBetween0to9 = Int.random(in: 0...9) 
let doubleBetween0to1 = Double.random(in: 0...1)

All number types have the static random(in:) function which takes the range and returns the random number in the given range.