java - 递增变量

Question

我理解后增量和前增量之间的区别，但我不明白这一点:

int b = 1;
b = b++;
System.out.println(b); // --> 1

为什么会打印 1？ b 获取 b++ 的值(即 1)后，b 不是递增吗？

int x = 1;
b = x++;
System.out.println(b); // --> 1
System.out.println(x); // --> 2

这是我期望的行为。两边使用相同的变量(b = b++)有什么不同？

Answer 1

如果你仔细观察字节码..你可以观察到差异..稍微改变你的代码..

public static void main(String[] args) {
    int b = 1;
    b = b++;
    System.out.println(b); // --> 1
    b=1;
    b=++b;
    System.out.println(b); // --> 2
}

字节码:

public static void main(java.lang.String[]);
   descriptor: ([Ljava/lang/String;)V
   flags: ACC_PUBLIC, ACC_STATIC
   Code:
     stack=2, locals=2, args_size=1
        0: iconst_1          --> int constant 1
        1: istore_1           --> store 1 in slot 1 of var table  i.e, for b  
        2: iload_1           --> load value of b from stack --> load 1
        3: iinc          1, 1  --> increment value b      --> store 2 for b 
        6: istore_1            --> store value of b "which was loaded"      --> store 1 to b.. So effectively 1 
        7: getstatic     #22                 // Field java/lang/System.out:Ljava
/io/PrintStream;
       10: iload_1
       11: invokevirtual #28                 // Method java/io/PrintStream.print
ln:(I)V
       14: iconst_1
       15: istore_1            --> store 1 in slot 1 of var table  i.e, for b
       16: iinc          1, 1  --> increment  --> b=2
       19: iload_1             --> load  2 
       20: istore_1            --> store 2.. So, effectively 2
       21: getstatic     #22                 // Field java/lang/System.out:Ljava
/io/PrintStream;
       24: iload_1
       25: invokevirtual #28                 // Method java/io/PrintStream.print
ln:(I)V
       28: return
     LineNumberTable:
       line 13: 0
       line 14: 2
       line 15: 7
       line 16: 14
       line 17: 16
       line 18: 21
       line 25: 28
     LocalVariableTable:
       Start  Length  Slot  Name   Signature
           0      29     0  args   [Ljava/lang/String;
           2      27     1     b   I