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假设你MethodHandle已经指定了一些参数,设置后如何更改这些参数?

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);
    }
}

我曾尝试使用 MethodHandles。过滤器参数()

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

    methodHandle.invoke(new SomeTest());
}

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

但我得到一个例外:

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)
4

1 回答 1

1

2种方法:

  • 重用您的原始方法句柄并将其绑定到另一个字符串:

    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"
    
  • 将第二个参数绑定到您操作的类成员的 getter。您必须使用“exactInvoker”过滤参数,该参数将执行 getter 以实际获取 String 值。看:

    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"
    }
    
于 2016-06-14T08:26:15.280 回答