Cod*_*Now 8 polymorphism inheritance scala dynamic-dispatch decompiler
反编译Scala代码:为什么派生类中有两个重写方法?
class A
{
private var str: String = "A"
val x: A = this
override def toString(): String = str
def m1(other: AnyRef): AnyRef = {
println("This is A.m1(AnyRef)")
other
}
}
class B extends A {
private var str: String = "B"
var z: Int = 0
override val x: B = this
override def m1(other: AnyRef): B = {
println("This is B.m1(AnyRef)")
this
}
}
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上面代码的B类反编译为:
public class test$B extends test$A {
private java.lang.String str;
private int z;
private final test$B x;
private java.lang.String str();
private void str_$eq(java.lang.String);
public int z();
public void z_$eq(int);
public test$B x();
public test$B m1(java.lang.Object);
public java.lang.Object m1(java.lang.Object);
public test$A x();
public test$B();
}
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我无法理解为什么m1反编译代码中有两个"版本"的方法.从我的理解,B.m1只是覆盖A.m1和public java.lang.Object m1(java.lang.Object)属于,A不应该在课堂上B.
这是一种合成桥接方法.
在Java字节码中,方法仅覆盖具有完全相同签名的方法.如果B没有任何实例Object m1(Object),那么任何调用它的尝试都会调用A中的实现,这不是你想要的.因此,编译器插入一个简单调用的合成桥接方法B m1(Object).此行为并非特定于Scala - 它也发生在纯Java中.
您可以通过检查反汇编来更详细地查看它.如果我编译并反汇编以下代码
class A
{
def m1(other: AnyRef): AnyRef = {
println("This is A.m1(AnyRef)")
other
}
}
class B extends A {
override def m1(other: AnyRef): B = {
println("This is B.m1(AnyRef)")
this
}
}
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B的相关部分是
.method public m1 : (Ljava/lang/Object;)LB;
.code stack 2 locals 2
L0: getstatic Field scala/Predef$ MODULE$ Lscala/Predef$;
L3: ldc 'This is B.m1(AnyRef)'
L5: invokevirtual Method scala/Predef$ println (Ljava/lang/Object;)V
L8: aload_0
L9: areturn
L10:
.end code
.methodparameters
other final
.end methodparameters
.end method
.method public bridge synthetic m1 : (Ljava/lang/Object;)Ljava/lang/Object;
.code stack 2 locals 2
L0: aload_0
L1: aload_1
L2: invokevirtual Method B m1 (Ljava/lang/Object;)LB;
L5: areturn
L6:
.end code
.methodparameters
other final
.end methodparameters
.end method
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如您所见,该方法m1 (Ljava/lang/Object;)Ljava/lang/Object;只是将参数转发给m1 (Ljava/lang/Object;)LB;.