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up vote 6 down vote favorite

This is a verification code generator, which generates n-digit numeric strings. (code can be 0000, so I chose String as a return type rather than Int or Long.)

It'll be great if anyone review this code and suggest more elegant or scalaish solution.

import scala.util.Random

object VerificationCodeGenerator {
  val rand = new Random

  def generate(digit: Int): String = {
    val sb = new StringBuilder
    for (i <- 1 to digit) {
      sb.append(rand.nextInt(10))
    }
    sb.toString()
  }
}
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2 Answers 2

up vote 7 down vote

if your number of digits is always going to be less than about 10, then you can use a single random operation and a string format, to do all the work without the loop.

Consider a method like:

def generate(digit: Int): String = {
    var randVal = rand.nextInt(math.pow(10, digit).toInt)
    var fmt = "%0" + digit + "d"
    fmt.format(randVal)
}

The randVal pulls a value with the limited number of digits (perhaps fewer than the limit). The format operation 0-pads the value to the right number, though.

This is not so much a scala way of doing it, but it is closer, and probably more efficient.

I would consider creating instances to handle each length of digits to avoid having to create the format each time, though.

See it running in ideone

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1  
In the general case, use either BigInteger or concatenate several generated strings. –  200_success♦ 22 hours ago
up vote 4 down vote

Shorter, more functional and perhaps more elegant way to do it is:

def generate(digits:Int):String = {
  (1 to digits) map(_=>rand.nextInt(10)) mkString("")
}

This code don't use mutable object(StringBuilder).

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