当超过 20 位的大整数时，为什么 Swift 5 String(Int) 会失败？

Question

我写了上面引用的简单代码来检查斐波那契数列中的整数是否不包含0或5，如果整数只包含1、2、3、4、6、7、8，则减少到1237，或 9 作为数字；如果是，则打印序列的成员。有趣的是，从数字游戏的角度来看，斐波那契数列中只有 23 个这样的整数。

当整数变大时，我必须使用 Swift-BigInt 库：

func getFib1237s() {
  // Some temporary variables.
  var a = BInt(0)
  var b = BInt(1)
  var m = BInt(1)
  var i = BInt(0)
  var z = BInt(1)
  // Get the numbers until crash...
  while i < z {
    let temp = a
    a = b
    b = b + temp
    print("a: ", a)
    var str = String(a)
    print("String start: ", str)
    str = str.replacingOccurrences(of: "9", with: "3")
    print("String after 9 reducto: ", str)
    str = str.replacingOccurrences(of: "6", with: "23")
    print("String after 6 reducto: ", str)
    str = str.replacingOccurrences(of: "8", with: "2")
    print("String after 8 reducto: ", str)
    str = str.replacingOccurrences(of: "4", with: "2")
    print("String after 4 reducto: ", str)
    if (str.firstIndex(of:"5") == nil) && (str.firstIndex(of: "0") == nil) && str.contains("1") && str.contains("2") && str.contains("3") && str.contains("7")  {
            print(m, "Fib 1237 number is ", a, " | Digits: ", str.count)
            m+=1
        }
    i+=1
    z+=1
  }
}

显然，在 20 位标记处或附近，String() 方法失败并抛出错误，不执行检查，因为根据调试器，整数被完全更改为随机的其他整数。

那么，在 Swift 中是否有任何 BigInt 或 String 解决方法/替代方案？我编写的 Ruby 代码在 Xcode 中运行良好，但我试图在这个项目中专门使用 Swift（和 metal），该项目最终需要在 iOS 上运行以用于商业/生产目的。

Answer 1

String(a) 调用采用BinaryInteger 的String.init 重载。这个初始化器很可能不是为处理超大数字而设计的。您可以改用a.asString(radix: 10) 转换为字符串。

为了使您的代码正常工作，您还应该：

• 删除(str.firstIndex(of: "0") == nil)
• 声明一个新的字符串变量并将替换后的字符串赋值给它，否则str.count会不正确。

我建议编写一个名为reduce 的单独方法，因为“减少”字符串需要很多步骤。

这里是reduce：

func reduce(_ s: String) -> String {
    let unique = String(Set(s))
    let replaced = unique.replacingOccurrences(of: "9", with: "3")
                                .replacingOccurrences(of: "6", with: "23")
                                .replacingOccurrences(of: "8", with: "2")
                                .replacingOccurrences(of: "4", with: "2")
                                .replacingOccurrences(of: "0", with: "")
    let sortedUniqueAgain = String(Set(replaced).sorted())
    return sortedUniqueAgain
}

现在，我们可以检查一下这个方法的返回值是否是1237：

while m <= 23 {
    let temp = a
    a = b
    b = b + temp
    let str = a.asString(radix: 10)
    // note that I have declared a new let constant here, instead of assigning to str
    // because otherwise str.count will be wrong
    let reduced = reduce(str)
    if reduced == "1237"  {
        print(m, "Fib 1237 number is ", a, " | Digits: ", str.count)
        m+=1
    }

}

输出：

1 Fib 1237 number is  317811  | Digits:  6
2 Fib 1237 number is  2178309  | Digits:  7
3 Fib 1237 number is  267914296  | Digits:  9
4 Fib 1237 number is  701408733  | Digits:  9
5 Fib 1237 number is  1134903170  | Digits:  10
6 Fib 1237 number is  72723460248141  | Digits:  14
7 Fib 1237 number is  117669030460994  | Digits:  15
8 Fib 1237 number is  8944394323791464  | Digits:  16
9 Fib 1237 number is  14472334024676221  | Digits:  17
10 Fib 1237 number is  37889062373143906  | Digits:  17
11 Fib 1237 number is  420196140727489673  | Digits:  18
12 Fib 1237 number is  1100087778366101931  | Digits:  19
13 Fib 1237 number is  1779979416004714189  | Digits:  19
14 Fib 1237 number is  2880067194370816120  | Digits:  19
15 Fib 1237 number is  19740274219868223167  | Digits:  20
16 Fib 1237 number is  83621143489848422977  | Digits:  20
17 Fib 1237 number is  927372692193078999176  | Digits:  21
18 Fib 1237 number is  781774079430987230203437  | Digits:  24
19 Fib 1237 number is  1264937032042997393488322  | Digits:  25
20 Fib 1237 number is  19134702400093278081449423917  | Digits:  29
21 Fib 1237 number is  1983924214061919432247806074196061  | Digits:  34
22 Fib 1237 number is  8404037832974134882743767626780173  | Digits:  34
23 Fib 1237 number is  162926777992448823780908130212788963731840407743629812913410  | Digits:  60