将Java 7编译为Java 6 [英] Compiling Java 7 to Java 6

问题描述

我知道Java 6不能使用Java 7的运行时功能，但是由于没有添加新的字节码，因此新的字节码invokedynamic仅与非Java相关.语言，我想知道将Java 7源代码(新的switch语句，diamond运算符)转换为纯Java 6有多困难(即能够在不失去Java 6兼容性的情况下开始将源转换为Java 7)

有指针吗?

解决方案

将Java 7 javac输出的.class文件标记为1.6.0版(即0x32)

printf "\x00\x00\x00\x32" |dd of=Example.class seek=4 bs=1 count=4 conv=notrunc

(根据 http://en.wikipedia.org/wiki/Java_class_file#General_layout )

如果将其(使用文件名使用$ 1)放入j6patch中，则可以使用以下方法来处理所有类文件:

find . -name \*.class |xargs -I {} ./j6patch {}

我在大型(〜4.8 MB jar)代码库上使用了此代码，甚至在Java 6 jar上使用了RetroTranslator，因此Java 7语言功能可以在以Java 5运行的应用程序上使用.Java 7编译器(javac)进行了很多额外的优化(例如转义分析)，从而显着提高了性能.

将RetroTranslator与-verify -target 1.5和JRE 1.6运行时jar一起使用，可以验证是否未使用Java 7运行时功能.

I'm aware that the runtime features of Java 7 are not available with Java 6 but since no new byte code has been added the new byte code invokedynamic is only relevant for non-Java languages, I was wondering how hard it would be to convert Java 7 source code (new switch statement, diamond operator) to pure Java 6 (i.e. to be able to start to convert the source to Java 7 without losing Java 6 compatibility).

Any pointers?

解决方案

Mark a .class file output by Java 7 javac with version 1.6.0 (i.e. 0x32)

printf "\x00\x00\x00\x32" |dd of=Example.class seek=4 bs=1 count=4 conv=notrunc

(according to http://en.wikipedia.org/wiki/Java_class_file#General_layout)

If you put that (using $1 for the filename) into j6patch you can do all class files with:

find . -name \*.class |xargs -I {} ./j6patch {}

I used this on a large (~4.8 MB jar) code base and even used RetroTranslator on the java 6 jar so Java 7 language features can be used on an app that runs in Java 5. Also the Java 7 compiler (javac) does lots of extra optimizations (e.g. escape analysis) that very noticeably improves performance.

Using RetroTranslator with -verify -target 1.5 and JRE 1.6 runtime jars allows to verify that no Java 7 runtime features are used.

这篇关于将Java 7编译为Java 6的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！