我无法理解ASMified Java字节码上的变量定位。我有以下Javacode:
public class TryCatch {
public static void main(String[] args) {
String test1 = null;
try {
String test2 ="try-inside-begin";
System.out.println("try-outside-begin");
try {
System.out.println(test2);
System.out.println(test1.length());
System.out.println("try-inside-end");
} catch (NullPointerException e) {
test2 = "catch-inside: " + e.getMessage();
throw new Exception(test2, e);
}
System.out.println("try-outside-end");
} catch (Exception e) {
System.out.println("catch-outside: " + e.getMessage());
} finally {
System.out.println("finally");
}
}
}
这成为main
的以下字节码:
TRYCATCHBLOCK L0 L1 L2 java/lang/NullPointerException
TRYCATCHBLOCK L3 L4 L5 java/lang/Exception
TRYCATCHBLOCK L3 L4 L6 null
TRYCATCHBLOCK L5 L7 L6 null
TRYCATCHBLOCK L6 L8 L6 null
L9
LINENUMBER 5 L9
ACONST_NULL
ASTORE 1
L3
LINENUMBER 7 L3
LDC "try-inside-begin"
ASTORE 2
L10
LINENUMBER 8 L10
GETSTATIC java/lang/System.out : Ljava/io/PrintStream;
LDC "try-outside-begin"
INVOKEVIRTUAL java/io/PrintStream.println (Ljava/lang/String;)V
L0
LINENUMBER 10 L0
GETSTATIC java/lang/System.out : Ljava/io/PrintStream;
ALOAD 2
INVOKEVIRTUAL java/io/PrintStream.println (Ljava/lang/String;)V
L11
LINENUMBER 11 L11
GETSTATIC java/lang/System.out : Ljava/io/PrintStream;
ALOAD 1
INVOKEVIRTUAL java/lang/String.length ()I
INVOKEVIRTUAL java/io/PrintStream.println (I)V
L12
LINENUMBER 12 L12
GETSTATIC java/lang/System.out : Ljava/io/PrintStream;
LDC "try-inside-end"
INVOKEVIRTUAL java/io/PrintStream.println (Ljava/lang/String;)V
L1
LINENUMBER 16 L1
GOTO L13
L2
LINENUMBER 13 L2
FRAME FULL [[Ljava/lang/String; java/lang/String java/lang/String] [java/lang/NullPointerException]
ASTORE 3
L14
LINENUMBER 14 L14
NEW java/lang/StringBuilder
DUP
INVOKESPECIAL java/lang/StringBuilder.<init> ()V
LDC "catch-inside: "
INVOKEVIRTUAL java/lang/StringBuilder.append (Ljava/lang/String;)Ljava/lang/StringBuilder;
ALOAD 3
INVOKEVIRTUAL java/lang/NullPointerException.getMessage ()Ljava/lang/String;
INVOKEVIRTUAL java/lang/StringBuilder.append (Ljava/lang/String;)Ljava/lang/StringBuilder;
INVOKEVIRTUAL java/lang/StringBuilder.toString ()Ljava/lang/String;
ASTORE 2
L15
LINENUMBER 15 L15
NEW java/lang/Exception
DUP
ALOAD 2
ALOAD 3
INVOKESPECIAL java/lang/Exception.<init> (Ljava/lang/String;Ljava/lang/Throwable;)V
ATHROW
L13
LINENUMBER 17 L13
FRAME SAME
GETSTATIC java/lang/System.out : Ljava/io/PrintStream;
LDC "try-outside-end"
INVOKEVIRTUAL java/io/PrintStream.println (Ljava/lang/String;)V
L4
LINENUMBER 21 L4
GETSTATIC java/lang/System.out : Ljava/io/PrintStream;
LDC "finally"
INVOKEVIRTUAL java/io/PrintStream.println (Ljava/lang/String;)V
L16
LINENUMBER 22 L16
GOTO L17
L5
LINENUMBER 18 L5
FRAME FULL [[Ljava/lang/String; java/lang/String] [java/lang/Exception]
ASTORE 2
L18
LINENUMBER 19 L18
GETSTATIC java/lang/System.out : Ljava/io/PrintStream;
NEW java/lang/StringBuilder
DUP
INVOKESPECIAL java/lang/StringBuilder.<init> ()V
LDC "catch-outside: "
INVOKEVIRTUAL java/lang/StringBuilder.append (Ljava/lang/String;)Ljava/lang/StringBuilder;
ALOAD 2
INVOKEVIRTUAL java/lang/Exception.getMessage ()Ljava/lang/String;
INVOKEVIRTUAL java/lang/StringBuilder.append (Ljava/lang/String;)Ljava/lang/StringBuilder;
INVOKEVIRTUAL java/lang/StringBuilder.toString ()Ljava/lang/String;
INVOKEVIRTUAL java/io/PrintStream.println (Ljava/lang/String;)V
L7
LINENUMBER 21 L7
GETSTATIC java/lang/System.out : Ljava/io/PrintStream;
LDC "finally"
INVOKEVIRTUAL java/io/PrintStream.println (Ljava/lang/String;)V
L19
LINENUMBER 22 L19
GOTO L17
L6
LINENUMBER 21 L6
FRAME SAME1 java/lang/Throwable
ASTORE 4
L8
GETSTATIC java/lang/System.out : Ljava/io/PrintStream;
LDC "finally"
INVOKEVIRTUAL java/io/PrintStream.println (Ljava/lang/String;)V
ALOAD 4
ATHROW
L17
LINENUMBER 23 L17
FRAME SAME
RETURN
MAXSTACK = 4
MAXLOCALS = 5
请注意底部附近有ASTORE 4
/ ALOAD 4
。为什么那是4而不是3?由于SAME1
框架是“与前一帧相同的局部区域,并且在堆栈上具有单个值”,因此前一帧只有两个局部区域(ref:FRAME FULL [[Ljava/lang/String; java/lang/String] [java/lang/Exception]
)。
我已阅读the spec,但我不清楚为什么它不是3。
答案 0 :(得分:2)
堆栈帧描述了局部变量的状态以及操作数堆栈出现的位置。后来的指令当然可以像平常那样修改。正确识别后,L6处的堆栈帧表示当控制流量达到L6时有两个局部变量。以下指令存储到插槽4,这是完全合法的。
了解堆栈映射的目的可能有所帮助。最初,根本没有堆栈映射,验证器使用推断来计算方法中每个点的局部变量。当遇到控制流时,它将合并该点的值并迭代直到收敛。
不幸的是,这很慢,所以为了加快速度,Oracle添加了堆栈映射。这基本上预先计算了加入控制流的任何点的验证结果。这样,验证程序可以通过代码执行单个线性传递,因为控制流不会更改结果。当验证者遇到控制流时,它会检查当前状态是否与跳转目标中声明的堆栈帧匹配,如果不匹配,则抛出错误。在线性代码的部分中,显然不需要包含堆栈帧,因为验证者可以执行与之前相同的操作。
堆栈帧不用于调试,它们旨在加快验证速度,因此它们包含验证所需的最少信息。如果编译器假设在每个指令处插入堆栈帧,那么astore 4
之后的堆栈帧当然会在第4个插槽中显示一个新变量。
至于为什么它在使用插槽3时使用插槽4,这只是编译器的一时兴起。也许它简化了javac的实现,但这只是推测。