在F#中從文本檔案中讀取輸入--用換行符讀取輸入的問題

這是來自 F# 中的 Project Euler #8 的代碼。

我的問題是我正在讀取一個由 \n 換行符分隔的大（1000 位）數字。但是，即使我在換行符上加入“字串”，我仍然會遇到錯誤。我已經以兩種不同的方式接近它，但我仍然收到相同的錯誤訊息。

open System;
open System.Text;

let path = "/Users/Arbin/Desktop/VS Code/F#/Project Euler/Largest Product in Series/largest_product_text_file"
//let monster_number = System.IO.File.ReadAllText path
let monster_number_array = System.IO.File.ReadAllLines path

let monster_number = String.Join("\n", System.IO.File.ReadAllLines path);


printfn "%s" monster_number

let adjacent n seq =
    seq |> Seq.mapi (fun index value -> seq |> Seq.skip index |> (Seq.truncate n))

//Mapping numbers from string to integer64.
let seq_of_seq = (adjacent 13 monster_number) |> Seq.map (Seq.map (int64 << string))

//Iterating through a sequence of sequence (nested sequence)
seq_of_seq |> Seq.iter (fun x -> x |> Seq.iter (fun y -> printfn "Ar: %A" y))

輸出：

<Sequence Integers>
System.FormatException: Input string was not in a correct format.
   at Microsoft.FSharp.Core.LanguagePrimitives.ParseInt64(String s) in D:\a\_work\1\s\src\FSharp.Core\prim-types.fs:line 2414
   at Microsoft.FSharp.Collections.Internal.IEnumerator.map@99.DoMoveNext(b& curr) in D:\a\_work\1\s\src\FSharp.Core\seq.fs:line 102
   at Microsoft.FSharp.Collections.Internal.IEnumerator.MapEnumerator`1.System.Collections.IEnumerator.MoveNext() in D:\a\_work\1\s\src\FSharp.Core\seq.fs:line 84
   at Microsoft.FSharp.Collections.SeqModule.Iterate[T](FSharpFunc`2 action, IEnumerable`1 source) in D:\a\_work\1\s\src\FSharp.Core\seq.fs:line 596
   at FSI_0001.it@20.Invoke(IEnumerable`1 x) in /Users/Arbin/Desktop/VS Code/F#/Project Euler/Largest Product in Series/largest_product_in_series.fsx:line 20
   at Microsoft.FSharp.Collections.SeqModule.Iterate[T](FSharpFunc`2 action, IEnumerable`1 source) in D:\a\_work\1\s\src\FSharp.Core\seq.fs:line 597
   at <StartupCode$FSI_0001>.$FSI_0001.main@() in /Users/Arbin/Desktop/VS Code/F#/Project Euler/Largest Product in Series/largest_product_in_series.fsx:line 20
Stopped due to error

號碼是：

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

uj5u.com熱心網友回復：

當您呼叫 時String.Join("\n", ...)，您將在結果字串中插入換行符。然后，稍后，您將呼叫int64 << string其中一個換行符。但是您不能將換行符轉換為整數。如果您在, 而不是"""中用作分隔符，則不會遇到此問題。（或者只是呼叫而不是。）String.Join"\nString.ConcatString.Join

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