Java ByteCode算术运算

本文介绍了Java ByteCode算术运算的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我要为学校项目制作一个简单的编译器，我想生成.class文件，我读取了文件格式，但是为了更好地理解.class文件格式和Java字节码，我有这个类:

I'm going to make a simple compiler for a school project, i want generate .class file, i read the file format but to understand better the .class file format and the java bytecode i have this class:

public class Me { public void myMethod() { int a = 5 * 4 + 3 - 2 + 1 / 7 + 28; } }

使用javap命令，我得到这个(用于"myMethod"):

with javap command i get this(for 'myMethod'):

public void myMethod(); flags: ACC_PUBLIC Code: stack=1, locals=2, args_size=1 0: bipush 49 2: istore_1 3: return LineNumberTable: line 3: 0 line 4: 3 LocalVariableTable: Start Length Slot Name Signature 0 4 0 this LMe; 3 1 1 a I

在这一行:

0: bipush 49

我不明白何时获得该数字(49)，我看不到算术运算'5 * 4 + 3 ...'的字节码

I dont understand when we get that number(49), i dont see the byte code for the arithmetic operation '5 * 4 + 3 ...'

推荐答案

我不知道何时获得该数字(49)，我看不到算术运算'5 * 4 + 3的字节码

I dont understand when we get that number(49), i dont see the byte code for the arithmetic operation '5 * 4 + 3

字节码编译器正在优化它们.根据JLS定义，该表达式是常量表达式，这意味着允许 java 编译器在编译时对其求值并将其硬编码为课程文件.

The bytecode compiler is optimizing them away. The expression is a constant expression according to the JLS definition, and that means that the java compiler is permitted to evaluate it at compile time and hard-code the resulting value into the class file.

如果要查看表达式评估的字节码是什么样的，则需要使用参数，局部变量等作为表达式中的主要"；例如

If you want to see what the bytecodes for expression evaluation look like, you need to use parameters, local variables, etc as the "primaries" in the expression; e.g.

public int myMethod(int a, int b, int c) { return a + b * c / 42; }