在查看下案例字节码时，对stack大小为何为4有过疑惑
public class TestClassFile {
public void add() {
int[] arr={1,2,3};
int i=arr[0]+arr[1];
}
}
字节码如下
descriptor: ()V
flags: ACC_PUBLIC
Code:
stack=4, locals=3, args_size=1
0: iconst_3
1: newarray int
3: dup
4: iconst_0
5: iconst_1
6: iastore
7: dup
8: iconst_1
9: iconst_2
10: iastore
11: dup
12: iconst_2
13: iconst_3
14: iastore
15: astore_1
16: aload_1
17: iconst_0
18: iaload
19: aload_1
20: iconst_1
21: iaload
22: iadd
23: istore_2
24: return
LineNumberTable:
line 5: 0
line 6: 16
line 7: 24
LocalVariableTable:
Start Length Slot Name Signature
0 25 0 this Lchapter_6/TestClassFile;
16 9 1 arr [I
24 1 2 i I
一开始以为是下方的加法运算造成的，但是分析后实际上加法运算只会使用两个栈空间，那么问题就出在一开始的数组定义上了。
实际上，数组在字节码中创建的过程如下：
- 数组长度入栈
- 数组指针入栈
- 数组index入栈
- 数组在该index下的值入栈
- 将值储存到数组index的位置，index与index下的值出栈，如果index<length，返回3；否则结束数组创建
在字节码中对应的是
descriptor: ()V
flags: ACC_PUBLIC
Code:
stack=4, locals=3, args_size=1
0: iconst_3 //数组长度
1: newarray int //创建数组
3: dup
4: iconst_0 //index=0
5: iconst_1 //arr[0]的值为1
6: iastore //储存
7: dup
8: iconst_1 //index=1
9: iconst_2 //arr[1]的值为2
10: iastore //储存
11: dup
12: iconst_2 //index=2
13: iconst_3 //arr[2]的值为3
14: iastore //储存
15: astore_1
16: aload_1
17: iconst_0
18: iaload
19: aload_1
20: iconst_1
21: iaload
22: iadd
23: istore_2
24: return
LineNumberTable:
line 5: 0
line 6: 16
line 7: 24
LocalVariableTable:
Start Length Slot Name Signature
0 25 0 this Lchapter_6/TestClassFile;
16 9 1 arr [I
24 1 2 i I
