2016-06-08 53 views
0

假設您有MethodHandle並且已經指定了一些參數,如何在設置後更改這些參數?如何在插入後更改MethodHandle參數?

import static java.lang.invoke.MethodType.*; 
import static java.lang.invoke.MethodHandles.*; 

import java.lang.invoke.MethodHandle; 
import java.lang.invoke.MethodHandles; 

public class SomeTest { 

    public static void main(String[] args) throws Throwable { 

     MethodHandle methodHandle = MethodHandles.lookup().findVirtual(SomeTest.class, 
       "someMethod", methodType(void.class, String.class)); 

     methodHandle = MethodHandles.insertArguments(methodHandle, 1, "Hi"); 

     // this invoke calls with "Hi", which is fine 
     methodHandle.invoke(new SomeTest()); 

     // here, how to change the arguments to be e.g. "Hello" instead of "Hi" 

     methodHandle.invoke(new SomeTest()); 

    } 

    public void someMethod(String a) { 
     System.out.println("Called with " + a); 
    } 
} 

我試過使用MethodHandles。 filterArguments()

.... 
    methodHandle = MethodHandles.filterArguments(methodHandle, 1, 
      MethodHandles.lookup().findStatic(SomeTest.class, "returnSomething", 
        methodType(String.class))); 

    methodHandle.invoke(new SomeTest()); 
} 

public static String returnSomething() { 
    return "Hello"; 
} 

,但我得到一個異常:

Exception in thread "main" java.lang.IllegalArgumentException: too many filters 
    at java.lang.invoke.MethodHandleStatics.newIllegalArgumentException(MethodHandleStatics.java:139) 
    at java.lang.invoke.MethodHandles.filterArgumentsCheckArity(MethodHandles.java:2623) 
    at java.lang.invoke.MethodHandles.filterArguments(MethodHandles.java:2595) 
    at test.test.SomeTest.main(SomeTest.java:22) 
+2

方法處理是不變的,*封裝*行爲。沒有辦法改變綁定值。如果要綁定不同的值,請保留具有該參數的原始方法句柄。 – Holger

回答

0

2種方法:

  • 重用原來的方法處理,並將其綁定到另一個字符串:

    MethodHandle methodHandle = MethodHandles.lookup().findVirtual(SomeTest.class, 
         "someMethod", methodType(void.class, String.class)); 
    MethodHandle hi = methodHandle.insertArguments(1, "Hi"); 
    MethodHandle hello = methodHandle.insertArguments(1, "Hello"); 
    hi.invoke(new SomeTest()); // "Hi" 
    hello.invoke(new SomeTest()); // "Hello" 
    
  • Bi找到你操作的類成員的getter的第二個參數。您必須使用「exactInvoker」過濾參數,該參數將執行getter以實際獲取String值。請參閱:

    public static class StringHolder{ 
        public String toPrint; 
        StringHolder(String toPrint){ 
         this.toPrint = toPrint; 
        } 
    } 
    
    public static void main(String[] args) throws Throwable { 
        MethodHandle toPrintGetter = lookup().findGetter(StringHolder.class, "toPrint", String.class); 
        MethodHandle someMethod = lookup().findVirtual(SomeTest.class, "someMethod", MethodType.methodType(void.class, String.class)); 
    
        StringHolder holder = new StringHolder("Hi"); 
        someMethod = MethodHandles.filterArguments(someMethod, 1, MethodHandles.exactInvoker(MethodType.methodType(String.class))); 
        MethodHandle stringPrinter = MethodHandles.insertArguments(someMethod, 1, toPrintGetter.bindTo(holder)); 
    
        stringPrinter.invokeExact(new SomeTest()); // prints "Hi" 
        holder.toPrint = "Hello"; 
        stringPrinter.invokeExact(new SomeTest()); // prints "Hello" 
    } 
    
+0

謝謝,但我得到'在線程中的異常「主要」java.lang.IllegalArgumentException:目標和過濾器類型不匹配:(SomeTest,字符串)無效,(MethodHandle)字符串' –

+0

是的,錯字在我身邊,索引filterArguments應該是1而不是0. – Gui13

+0

謝謝,但答案會在第一次調用之前更改方法句柄。請檢查我發佈的代碼,我需要在設置參數後更改methodHandle。 –

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