浅谈JVM 底层解析 i++和 ++i 区别-编程学习网

一、前言

如果只用普通的知识解释i++和++i的话
i++ 先将i赋值再++
++i 先++再赋值
但是这简单的回答并不能入吸引面试官的眼球,如果用java字节码指令分析则效果完全不同

二、代码实现


public class OperandStackTest {

    public static void add(){
        //第1类问题：
        int i1 = 10;
        i1++;
        System.out.println("i1 =" + i1);//11

        int i2 = 10;
        ++i2;
        System.out.println("i2 =" + i2);//11

        //第2类问题：
        int i3 = 10;
        int i4 = i3++;
        System.out.println("i3 =" + i3);//11
        System.out.println("i4 =" + i4);//10

        int i5 = 10;
        int i6 = ++i5;
        System.out.println("i5 =" + i5);//11
        System.out.println("i6 =" + i6);//11

        //第3类问题：
        int i7 = 10;
        i7 = i7++;
        System.out.println("i7 =" + i7);//10

        int i8 = 10;
        i8 = ++i8;
        System.out.println("i8 =" + i8);//11

        //第4类问题：
        int i9 = 10;
        int i10 = i9++ + ++i9;//10+12
        System.out.println("i9 =" + i9);//12
        System.out.println("i10 =" + i10);//22
    }

    public static void main(String[] args) {
        add();
    }
}

运行结果


i1 = 11
i2 = 11
i3 = 11
i4 = 10
i5 = 11
i6 = 11
i7 = 10
i8 = 11
i9 = 12
i10 = 22

三、字节码指令

通过javap -v out目录下的class文件名在终端运行得到如下结果


public static void add();
    descriptor: ()V
    flags: ACC_PUBLIC, ACC_STATIC
    Code:
      stack=2, locals=10, args_size=0
         0: bipush        10
         2: istore_0
         3: iinc          0, 1
         6: getstatic     #4                  // Field java/lang/System.out:Ljava/io/PrintStream;
         9: iload_0
        10: invokevirtual #5                  // Method java/io/PrintStream.println:(I)V
        13: bipush        10
        15: istore_1
        16: iinc          1, 1
        19: getstatic     #4                  // Field java/lang/System.out:Ljava/io/PrintStream;
        22: iload_1
        23: invokevirtual #5                  // Method java/io/PrintStream.println:(I)V
        26: bipush        10
        28: istore_2
        29: iload_2
        30: iinc          2, 1
        33: istore_3
        34: getstatic     #4                  // Field java/lang/System.out:Ljava/io/PrintStream;
        37: iload_2
        38: invokevirtual #5                  // Method java/io/PrintStream.println:(I)V
        41: getstatic     #4                  // Field java/lang/System.out:Ljava/io/PrintStream;
        44: iload_3
        45: invokevirtual #5                  // Method java/io/PrintStream.println:(I)V
        48: bipush        10
        50: istore        4
        52: iinc          4, 1
        55: iload         4
        57: istore        5
        59: getstatic     #4                  // Field java/lang/System.out:Ljava/io/PrintStream;
        62: iload         4
        64: invokevirtual #5                  // Method java/io/PrintStream.println:(I)V
        67: getstatic     #4                  // Field java/lang/System.out:Ljava/io/PrintStream;
        70: iload         5
        72: invokevirtual #5                  // Method java/io/PrintStream.println:(I)V
        75: bipush        10
        77: istore        6
        79: iload         6
        81: iinc          6, 1
        84: istore        6
        86: getstatic     #4                  // Field java/lang/System.out:Ljava/io/PrintStream;
        89: iload         6
        91: invokevirtual #5                  // Method java/io/PrintStream.println:(I)V
        94: bipush        10
        96: istore        7
        98: iinc          7, 1
       101: iload         7
       103: istore        7
       105: getstatic     #4                  // Field java/lang/System.out:Ljava/io/PrintStream;
       108: iload         7
       110: invokevirtual #5                  // Method java/io/PrintStream.println:(I)V
       113: bipush        10
       115: istore        8
       117: iload         8
       119: iinc          8, 1
       122: iinc          8, 1
       125: iload         8
       127: iadd
       128: istore        9
       130: getstatic     #4                  // Field java/lang/System.out:Ljava/io/PrintStream;
       133: iload         8
       135: invokevirtual #5                  // Method java/io/PrintStream.println:(I)V
       138: getstatic     #4                  // Field java/lang/System.out:Ljava/io/PrintStream;
       141: iload         9
       143: invokevirtual #5                  // Method java/io/PrintStream.println:(I)V
       146: return

四、字节码解析

1. 第一类问题


 //第1类问题：
 int i1 = 10;
 i1++;
 System.out.println("i1 =" + i1);//11

 int i2 = 10;
 ++i2;
 System.out.println("i2 =" + i2);//11

对应字节码指令为


   0: bipush        10
   2: istore_0
   3: iinc          0, 1
   6: getstatic     #4                  // Field java/lang/System.out:Ljava/io/PrintStream;
   9: iload_0
  10: invokevirtual #5                  // Method java/io/PrintStream.println:(I)V
  13: bipush        10
  15: istore_1
  16: iinc          1, 1
  19: getstatic     #4                  // Field java/lang/System.out:Ljava/io/PrintStream;
  22: iload_1
  23: invokevirtual #5                  // Method java/io/PrintStream.println:(I)V

先将i1的值为10入栈(bipush),然后将int类型的值从栈中存到局部变量表0的位置,然后执行iinc将0位置的值+1,然后将局部变量表0位置的数入栈执行输出操作

所以i1的值为11

先将i2的值为10入栈(bipush),然后将int类型的值从栈中存到局部变量表1的位置,然后执行iinc将1位置的值+1,然后将局部变量表1位置的数入栈执行输出操作

所以i2的值为11

由于没有赋值操作,区别不大

2. 第二类问题


//第2类问题：
 int i3 = 10;
 int i4 = i3++;
 System.out.println("i3 =" + i3);//11
 System.out.println("i4 =" + i4);//10

 int i5 = 10;
 int i6 = ++i5;
 System.out.println("i5 =" + i5);//11
 System.out.println("i6 =" + i6);//11

对应字节码为


  26: bipush        10
  28: istore_2
  29: iload_2
  30: iinc          2, 1
  33: istore_3
  34: getstatic     #4                  // Field java/lang/System.out:Ljava/io/PrintStream;
  37: iload_2
  38: invokevirtual #5                  // Method java/io/PrintStream.println:(I)V
  41: getstatic     #4                  // Field java/lang/System.out:Ljava/io/PrintStream;
  44: iload_3
  45: invokevirtual #5                  // Method java/io/PrintStream.println:(I)V
  48: bipush        10
  50: istore        4
  52: iinc          4, 1
  55: iload         4
  57: istore        5
  59: getstatic     #4                  // Field java/lang/System.out:Ljava/io/PrintStream;
  62: iload         4
  64: invokevirtual #5                  // Method java/io/PrintStream.println:(I)V
  67: getstatic     #4                  // Field java/lang/System.out:Ljava/io/PrintStream;
  70: iload         5
  72: invokevirtual #5                  // Method java/io/PrintStream.println:(I)V

先将i3入栈存储到局部变量表2的位置,然后将它入栈,执行iinc将2位置的值加一,i4存储到局部表量表3的位置

所以i3是11,i4还是10

将i5入栈存储到局部变量表4的位置,由于是++i所以先iinc将4位置的值加一,然后将局部变量表4的值入栈,执行赋值操作

所以i5、i6都是11

3. 第三类问题


 //第3类问题：
 int i7 = 10;
 i7 = i7++;
 System.out.println("i7 =" + i7);//10
 int i8 = 10;
 i8 = ++i8;
 System.out.println("i8 =" + i8);//11

对应字节码


  75: bipush        10
  77: istore        6
  79: iload         6
  81: iinc          6, 1
  84: istore        6
  86: getstatic     #4                  // Field java/lang/System.out:Ljava/io/PrintStream;
  89: iload         6
  91: invokevirtual #5                  // Method java/io/PrintStream.println:(I)V
  94: bipush        10
  96: istore        7
  98: iinc          7, 1
 101: iload         7
 103: istore        7
 105: getstatic     #4                  // Field java/lang/System.out:Ljava/io/PrintStream;
 108: iload         7
 110: invokevirtual #5                  // Method java/io/PrintStream.println:(I)V

先将i7入栈,然后存到局部变量表6的位置,先把i6入栈,然后把6处的值加一,由于又将这个值存储到局部变量表6处,所以产生覆盖又把值变为10

所以i7为10

而++i不会产生覆盖先执行加一然后再把值入栈,在赋值给局部变量表中

所以i8为11

4. 第四类问题


 //第4类问题：
 int i9 = 10;
 int i10 = i9++ + ++i9;//10+12
 System.out.println("i9 =" + i9);//12
 System.out.println("i10 =" + i10);//22

对应字节码为


 113: bipush        10
 115: istore        8
 117: iload         8
 119: iinc          8, 1
 122: iinc          8, 1
 125: iload         8
 127: iadd
 128: istore        9
 130: getstatic     #4                  // Field java/lang/System.out:Ljava/io/PrintStream;
 133: iload         8
 135: invokevirtual #5                  // Method java/io/PrintStream.println:(I)V
 138: getstatic     #4                  // Field java/lang/System.out:Ljava/io/PrintStream;
 141: iload         9
 143: invokevirtual #5                  // Method java/io/PrintStream.println:(I)V
 146: return

先将i9=10入栈,然后存在局部变量表8的位置
int i10 = i9++ + ++i9;
先iload将8位置的i9入栈然后执行iinc将8处的i9加一,然后执行++i9,在将8处的i9加一

此时i9=10+1+1为12

然后将8位置的i9入栈,执行add将栈中的两i9相加,得到的值存储到局部变量表9的位置

所以i10=10+12(i9++后还是10,++i9后是12,因为执行了两次iinc操作)

然后调用虚方法和静态方法,在将9处的值入栈执行输出语句

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