最后了解 JVM 字节码中的本地 var 位置
Understanding local var position in JVM bytecode on finally
我无法理解 ASMified Java 字节码上的变量定位。我有以下 Java 代码:
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");
}
}
}
它成为 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 ASTORE 4
/ALOAD 4
离底部有多近。为什么是 4 而不是 3?由于 SAME1
帧是 "same locals as the previous frame and with a single value on the stack" 并且前一帧只有两个局部变量(参考:FRAME FULL [[Ljava/lang/String; java/lang/String] [java/lang/Exception]
)。
我已经阅读了 the spec,但我也不清楚为什么它不是 3。
栈帧描述局部变量的状态和操作数栈在它出现的地方。后面的指令当然可以修改正常的东西。正如您正确识别的那样,L6 的堆栈帧表示当控制流到达 L6 时有两个局部变量。下面指令存储到slot 4,完全合法
这可能有助于理解堆栈映射的用途。最初,根本没有堆栈映射,验证者使用推理来计算方法中每个点的局部变量。当遇到控制流时,它会合并那个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。这实质上是在加入控制流的任何点预先计算验证结果。这样,验证者可以对代码进行一次线性传递,因为控制流不会改变结果。当验证器遇到控制流时,它检查当前状态是否与在跳转目标处声明的堆栈帧匹配,如果不匹配,则抛出错误。在线性代码段中,显然没有必要包含堆栈帧,因为验证器可以做与之前相同的事情。
堆栈帧不是用来调试的,它们是用来加速验证的,所以它们包含验证所需的最少信息。如果编译器假设在 每 指令处插入一个堆栈帧,那么 astore 4
之后的堆栈帧当然会在第 4 个槽中显示一个新变量。
至于为什么它可以使用插槽 3 而使用插槽 4,那只是编译器的心血来潮。也许它简化了 javac 的实现,但这只是推测。
我无法理解 ASMified Java 字节码上的变量定位。我有以下 Java 代码:
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");
}
}
}
它成为 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 ASTORE 4
/ALOAD 4
离底部有多近。为什么是 4 而不是 3?由于 SAME1
帧是 "same locals as the previous frame and with a single value on the stack" 并且前一帧只有两个局部变量(参考:FRAME FULL [[Ljava/lang/String; java/lang/String] [java/lang/Exception]
)。
我已经阅读了 the spec,但我也不清楚为什么它不是 3。
栈帧描述局部变量的状态和操作数栈在它出现的地方。后面的指令当然可以修改正常的东西。正如您正确识别的那样,L6 的堆栈帧表示当控制流到达 L6 时有两个局部变量。下面指令存储到slot 4,完全合法
这可能有助于理解堆栈映射的用途。最初,根本没有堆栈映射,验证者使用推理来计算方法中每个点的局部变量。当遇到控制流时,它会合并那个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。这实质上是在加入控制流的任何点预先计算验证结果。这样,验证者可以对代码进行一次线性传递,因为控制流不会改变结果。当验证器遇到控制流时,它检查当前状态是否与在跳转目标处声明的堆栈帧匹配,如果不匹配,则抛出错误。在线性代码段中,显然没有必要包含堆栈帧,因为验证器可以做与之前相同的事情。
堆栈帧不是用来调试的,它们是用来加速验证的,所以它们包含验证所需的最少信息。如果编译器假设在 每 指令处插入一个堆栈帧,那么 astore 4
之后的堆栈帧当然会在第 4 个槽中显示一个新变量。
至于为什么它可以使用插槽 3 而使用插槽 4,那只是编译器的心血来潮。也许它简化了 javac 的实现,但这只是推测。