生成一定位数的随机数

Generate random number of certain amount of digits(生成一定位数的随机数)

问题描述

喂，

我有一个非常基本的问题:

I have a very Basic Question which is :

如何在 Swift 中创建一个 20 位无浮点数无负数(基本上是 Int)的随机数?

How can i create a random number with 20 digits no floats no negatives (basically an Int) in Swift ?

感谢所有回答XD

推荐答案

这里有一些伪代码可以满足你的需求.

Here is some pseudocode that should do what you want.

generateRandomNumber(20)
func generateRandomNumber(int numDigits){
   var place = 1
   var finalNumber = 0;
   for(int i = 0; i < numDigits; i++){
      place *= 10
      var randomNumber = arc4random_uniform(10)
      finalNumber += randomNumber * place
  }
  return finalNumber
}

它非常简单.你生成 20 个随机数，然后将它们乘以它们应该在的十位、百分之一、千位…….这样，您将保证数字的大小正确，但会随机生成将在每个地方使用的数字.

Its pretty simple. You generate 20 random numbers, and multiply them by the respective tens, hundredths, thousands... place that they should be on. This way you will guarantee a number of the correct size, but will randomly generate the number that will be used in each place.

更新

正如评论中所说，您很可能会遇到一个带有这么长数字的溢出异常，因此您必须创造性地存储数字(字符串等)，但我只是想向您展示一种生成具有保证数字长度的数字的简单方法.此外，鉴于当前代码，您的前导数字可能为 0 的可能性很小，因此您也应该防止这种情况发生.

As said in the comments you will most likely get an overflow exception with a number this long, so you'll have to be creative in how you'd like to store the number (String, ect...) but I merely wanted to show you a simple way to generate a number with a guaranteed digit length. Also, given the current code there is a small chance your leading number could be 0 so you should protect against that as well.

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