为什么Swift 5 String（Int）的整数大于20位时失败？

Question

我编写了上面引用的简单代码，以检查斐波那契数列中的整数是否不包含0或5，如果整数仅包含1,2,3,4,6,7,8，则减少到1237。 9位数字；如果是这样，则打印该序列的成员。从数字游戏的角度来看，有趣的是，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()方法失败并引发错误，没有执行检查，因为根据调试器，该整数将完全更改为随机的其他整数。

因此，在Swift中是否有BigInt或String解决方法/替代方法？我编写的Ruby代码在Xcode中可以正常工作，但是我试图为此项目专门使用Swift（和金属），最终需要在iOS上进行商业/生产用途。

Answer 1

String(a)调用需要String.init的BinaryInteger重载。此初始化程序很可能没有设计为处理超大数字。您可以使用a.asString(radix: 10)转换为字符串。

为了使代码正常工作，您还应该：

• 删除(str.firstIndex(of: "0") == nil)
• 声明一个新的字符串变量并为其分配替换的字符串，否则str.count将不正确。

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，我们只需检查此方法的返回值是否为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