Array.find（）提供了奇怪的结果

Question

我正在为我的兼职编程课程写作业。我的代码的问题是array.find（）和搜索的结果。它应该（在我的理论中）在数组中搜索信息，然后将它们发布给用户，但是所有搜索结果都是一样的：ass2task1.Program + customer这里只有代码的一部分，因为老师告诉我们，我们可以在互联网上，只要张贴问题，因为我们当您转换对象为字符串（"The forename resuts:" + resultforename）不要张贴我们的整个代码

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()将返回匹配的断言，如果你想要的属性值，你需要做的是这样的对象：resultforename.customerforname或类似的东西。

如果没有找到它，那么默认值将被退回，所以检查空值等

Answer 2

它调用对象的方法ToString()。定义一个适当的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

要在里克和clcto的回答（尝试）扩展。在

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

你resultforename你得到的结构名称的原因是客户结构的 - 默认Console.WriteLine（结构）不知道如何表达一个复杂的对象为字符串。

至于建议你可以做

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

或为结构提供自己的ToString（）方法clcto指出的 - 这样做基本上就Console.WriteLine（或任何字符串表示）如何将客户的结构表示为字符串。

不知道这是否会有所帮助，或使其更加清晰。但给定：

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"