我的陈述是真的吗?
这是我的密码:
public void start() {
Consumer<Integer> someFunc = (someInt) -> {
System.out.println("Hello lambda!");
};
}还有我的代码的字节码:
~启动法
// access flags 0x1
public start()V
L0
LINENUMBER 9 L0
INVOKEDYNAMIC accept()Ljava/util/function/Consumer; [
// handle kind 0x6 : INVOKESTATIC
java/lang/invoke/LambdaMetafactory.metafactory(Ljava/lang/invoke/MethodHandles$Lookup;Ljava/lang/String;Ljava/lang/invoke/MethodType;Ljava/lang/invoke/MethodType;Ljava/lang/invoke/MethodHandle;Ljava/lang/invoke/MethodType;)Ljava/lang/invoke/CallSite;
// arguments:
(Ljava/lang/Object;)V,
// handle kind 0x6 : INVOKESTATIC
me/alexandr/SomeMainClass.lambda$start$0(Ljava/lang/Integer;)V,
(Ljava/lang/Integer;)V
]
ASTORE 1
L1
LINENUMBER 12 L1
ALOAD 1
ICONST_1
INVOKESTATIC java/lang/Integer.valueOf (I)Ljava/lang/Integer;
INVOKEINTERFACE java/util/function/Consumer.accept (Ljava/lang/Object;)V
L2
LINENUMBER 13 L2
RETURN
L3
LOCALVARIABLE this Lme/alexandr/SomeMainClass; L0 L3 0
LOCALVARIABLE someFunc Ljava/util/function/Consumer; L1 L3 1
// signature Ljava/util/function/Consumer<Ljava/lang/Integer;>;
// declaration: java.util.function.Consumer<java.lang.Integer>
MAXSTACK = 2
MAXLOCALS = 2~翻译lambda表达式
// access flags 0x100A
private static synthetic lambda$start$0(Ljava/lang/Integer;)V
L0
LINENUMBER 10 L0
GETSTATIC java/lang/System.out : Ljava/io/PrintStream;
LDC "Hello lambda!"
INVOKEVIRTUAL java/io/PrintStream.println (Ljava/lang/String;)V
L1
LINENUMBER 11 L1
RETURN
L2
LOCALVARIABLE someInt Ljava/lang/Integer; L0 L2 0
MAXSTACK = 2
MAXLOCALS = 1据我所见-一个lambda表达式被翻译成一个静态方法。那么,我可以说函数接口是静态方法的“杠杆”,允许我调用它吗?
发布于 2015-08-13 21:23:48
https://stackoverflow.com/questions/31997609
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