Question

我需要将Java代码转换为Scala，但是编译器向我显示错误。我知道方法上的参数输入是val类型。如果我需要转换这些值，我可以采用哪种选择？我想应用案例类或类...在代码段代码下方（在Scala中）：

def pmerge_FA(x: Pennant,y: Pennant): Pennant={

        if(x == null && y == null && this.root == null){
            return null
        }else if(x == null && y == null){
            return this
        }else if(this.root == null && x == null){
            return y
        }else if(this.root == null && y == null){
            return x
        }else if(x == null){
            y = y.pmerge(this) //error
                    null
        }else if(this.root == null){
            y = y.pmerge(x) //error
                    null
        }else if (y == null){
            y = this.pmerge(x) // error
                    null
        }else{
            y = y.pmerge(x)
                    this
        }
}

请注意，更新y参数时显示错误。

谢谢

Answer 1

是的，由于无法将某些内容重新分配给val以及Scala中方法的参数仅作为val（不可变的）发送，因此显示了错误。

由于您没有提供'Public Declare Sub Sleep Lib "kernel32" (ByVal dwMilliseconds As Long) Sub login() Const Url$ = "https://www.hedgefundresearch.com/user/login" Dim oLogin As Object, oPassword As Object, oButton As Object, HTMLdoc As Object Dim UserName As String, Password As String UserName = "yyyyy@zzzz.com" Password = "password" Dim ie As Object, myURL As String Set ie = CreateObject("InternetExplorer.Application") ie.navigate Url ie.Visible = True Do Until ie.ReadyState = 4 DoEvents Loop Set HTMLdoc = ie.document Set oLogin = HTMLdoc.getElementsByClassName("form-control form-text required")(0) Set oPassword = HTMLdoc.getElementsByClassName("form-control form-text required")(1) Set oButton = HTMLdoc.getElementsByClassName("btn btn-primary form-submit icon-before")(0) On Error Resume Next oLogin.Value = UserName On Error Resume Next oPassword.Value = Password On Error Resume Next oButton.Click myURL = "https://www.hedgefundresearch.com/download/index-ror-perf-download/2899" ie.navigate myURL 'AppActivate "zzz@yyyy.net | Hedge Fund Research® - Internet Explorer" 'Sleep 1 Application.SendKeys ("%s") End Sub的完整定义，因此很难提出替代解决方案，但是：

通常，在Scala中，您可以使用pattern matching来代替GET /download/index-ror-perf-download/2899 HTTP/1.1 Host: www.hedgefundresearch.com Connection: keep-alive Upgrade-Insecure-Requests: 1 User-Agent: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/73.0.3683.86 Safari/537.36 Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,image/apng,*/*;q=0.8,application/signed-exchange;v=b3 Referer: https://www.hedgefundresearch.com/indices/hfri-fund-weighted-composite-index Accept-Encoding: gzip, deflate, br Accept-Language: en-US,en;q=0.9 Cookie: _ga=GA1.2.1864475516.1551721637; COOKIE_TOU=2019-03-04+11%3A47%3A19; cookiesession1=297E7ACEBCDZVZC3YAZR1VC8IWL6AF6D; has_js=1; _gid=GA1.2.2018233246.1554731723; SSESScd329e696f514d42291a588234027dae=G6jjp0oCZUb2FjW43IXgTVZAOsb2UbyoU7iawYa2erw General: Request URL: https://www.hedgefundresearch.com/download/index-ror-perf-download/2899 Request Method: GET Status Code: 200 OK Remote Address: 167.88.151.170:443 Referrer Policy: no-referrer-when-downgrade“ Java”样式，而可以使用Option来代替this，这非常强大

例如，我建议以这种“ Scala”样式（部分实现）重构您的方法

if-else

这样一来，您将返回null，def pmerge_FA(x: Pennant, y: Pennant): Option[Pennant] = { (Option(x),Option(y), Option(this.root)) match { case (None, None, None) => None case (None, None, _) => Option("") case (None, _, None) => Option(y) case (_, None, None) => Option(x) case (None, _, _) => .... } }作为它们的新值，或者创建一个case类，例如：

并在需要时将其返回。同样，如果您将提供有关y类的更多信息，则可以为该方法提供更好的替代实现。

Answer 2

分配后再也不会使用y（即y.pmerge(...)）的新值。因此，我猜所有的赋值y = y.pmerge(...)都可以被调用y.pmerge(...)代替。

y.pmerge(...)有副作用吗？以防万一，如果没有，则永远不会使用值y.pmerge(...)（仅返回null或this），因此在这种情况下，可以完全删除行y = y.pmerge(...)。 / p>

所以代码可以是（如果有副作用）

def pmerge_FA(x: Pennant,y: Pennant): Pennant={

        if(x == null && y == null && this.root == null){
            null
        }else if(x == null && y == null){
            this
        }else if(this.root == null && x == null){
            y
        }else if(this.root == null && y == null){
            x
        }else if(x == null){
            y.pmerge(this)
            null
        }else if(this.root == null){
            y.pmerge(x)
            null
        }else if (y == null){
            this.pmerge(x)
            null
        }else{
            y.pmerge(x)
            this
        }
}

或（如果没有副作用）

def pmerge_FA(x: Pennant,y: Pennant): Pennant={

        if(x == null && y == null && this.root == null){
            null
        }else if(x == null && y == null){
            this
        }else if(this.root == null && x == null){
            y
        }else if(this.root == null && y == null){
            x
        }else if(x == null){
            null
        }else if(this.root == null){
            null
        }else if (y == null){
            null
        }else{
            this
        }
}

Answer 3

哦，对了！共有三个类来构建Bag数据结构对象，并在完全平衡的树中添加节点。这些方法适用于该方法。下面是完整的代码（使用Java）。 Pennant类使用图对象的节点构建森林。

节点类：

公共类节点{

==

}

公共类锦旗{

    private Node left;
    private Node right;
    private int item;

    public Node() {
        left = null;
        right = null;
        item = 0;
    }
    public Node(int value) {
        left = null;
        right = null;
        item = value;
    }

    public Node getLeft() {
        return left;
    }
    public void setLeft(Node left) {
        this.left = left;
    }
    public Node getRight() {
        return right;
    }
    public void setRight(Node right) {
        this.right = right;
    }
    public int getItem() {
        return this.item;
    }
    public void setItem(int item) {
        this.item = item;
    }

}

在Scala中传递方法的参数（更改变量值的替代方法）

3 个答案: