Array.find（）提供了奇怪的結果

Question

我正在爲我的兼職編程課程寫作業。我的代碼的問題是array.find（）和搜索的結果。它應該（在我的理論中）在數組中搜索信息，然後將它們發佈給用戶，但是所有搜索結果都是一樣的：ass2task1.Program + customer這裏只有代碼的一部分，因爲老師告訴我們，我們可以在互聯網上，只要張貼問題，因爲我們當您轉換對象爲字符串（"The forename resuts:" + resultforename）不要張貼我們的整個代碼

struct customer 
    { 
     public int customernumber; 
     public string customersurname; 
     public string customerforname; 
     public string customerstreet; 
     public string customertown; 
     public DateTime customerdob; 
    } 

    static void Main(string[] args) 
    { 
     customer[] customerdetails = new customer[99]; 
     int selector = 0; 
     int selector2 = 0; 
     string vtemp = ""; 
     string ctemp = ""; 
     int searchnumber; 
     string searchforename; //variable/ array declaring 
     string searchsurname; 
     string searchtown; 
     DateTime searchdob; 
     customer resultnumber; 
     customer resultforename; 
     customer resultsurname; 
     customer resulttown; 
     customer resultdob; 


    if (selector2 == 2) 
        { 
         Console.Clear(); 
         Console.WriteLine("Enter the forename you are looking for: "); 
         searchforename = (Console.ReadLine()); 
         resultforename = Array.Find(customerdetails, customer => customer.customerforname == searchforename); 
         Console.Clear(); 
         Console.WriteLine("Enter the surname you are looking for: "); // all of the searches comes out with ass2task1.Program+customer result 
         searchsurname = (Console.ReadLine()); 
         resultsurname = Array.Find(customerdetails, customer => customer.customersurname == searchsurname); 
         Console.WriteLine("The forename resuts:" + resultforename); 
         Console.WriteLine("The surname resuts:" + resultsurname); 

Answer 1

Array.Find()將返回匹配的斷言，如果你想要的屬性值，你需要做的是這樣的對象：resultforename.customerforname或類似的東西。

如果沒有找到它，那麼默認值將被退回，所以檢查空值等

Answer 2

它調用對象的方法ToString()。定義一個適當的ToString()方法：

struct customer 
{ 
    public int customernumber; 
    public string customersurname; 
    public string customerforname; 
    public string customerstreet; 
    public string customertown; 
    public DateTime customerdob; 

    public override string ToString() 
    { 
     return customersurname + ", " + customerforname; 
    } 
} 

Answer 3

要在裏克和clcto的回答（嘗試）擴展。在

Console.WriteLine("The forename resuts:" + resultforename); 
Console.WriteLine("The surname resuts:" + resultsurname); 

你resultforename你得到的結構名稱的原因是客戶結構的 - 默認Console.WriteLine（結構）不知道如何表達一個複雜的對象爲字符串。

至於建議你可以做

Console.WriteLine("The forename resuts:" + resultforename.customerforname); 

或爲結構提供自己的ToString（）方法clcto指出的 - 這樣做基本上就Console.WriteLine（或任何字符串表示）如何將客戶的結構表示爲字符串。

不知道這是否會有所幫助，或使其更加清晰。但給定：

public struct Foo 
{ 
    public string Bar { get; set; } 
} 

public struct FooTwo 
{ 
    public string Bar { get; set; } 

    public override string ToString() 
    { 
     return "This is how to represent a Foo2 as string: " + Bar; 
    } 
} 

Foo[] foos = new Foo[99]; 

Foo foundFoo = foos[0]; // This is equivalent to your find statement... setting a foundFoo local variable to a Foo struct 
string foundBar = foos[0].Bar; // This is another way to get to what you're trying to accoomplish, setting a string value representation of your found object. 

Console.WriteLine(foundFoo); // Doesn't know how to deal with printing out a Foo struct - simply writes [namespace].Foo 
Console.WriteLine(foundFoo.Bar); // Outputs Foo.Bar value 
Console.WriteLine(foundBar); // Outputs Foo.Bar value 

FooTwo foo2 = new FooTwo(); 
foo2.Bar = "test bar"; 

Console.WriteLine(foo2); // outputs: "This is how to represent a Foo2 as string: test bar"