preface
If I ++ and ++ I were just general knowledge
- I ++, I assign I and then I ++
- ++ I ++ and then assign
But this simple answer won’t catch the interviewer’s eye, and when analyzed using Java bytecode instructions, it will.
Code implementation
Public class OperandStackTest {/** public static void add(){// public static void add(){// public static void add(); i1++; System.out.println(i1); //11 int i2 = 10; ++i2; System.out.println(i2); // int i3 = 10; int i4 = i3++; System.out.println(i3); //11 System.out.println(i4); //10 int i5 = 10; int i6 = ++i5; System.out.println(i5); //11 System.out.println(i6); Int i7 = 10; // int i7 = 10; i7 = i7++; System.out.println(i7); //10 int i8 = 10; i8 = ++i8; System.out.println(i8); // int i9 = 10; int i10 = i9++ + ++i9; //10+12 System.out.println(i9); //12 System.out.println(i10); //22 } public static void main(String[] args) { add(); }}Copy the codeThe results
Bytecode instruction
Run the class file name in the javap -v out directory on the terminal to obtain the following result
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
Copy the codeExplain the above results
The first kind of question
The corresponding instruction is
I1 is pushed with a value of 10 (bipush), then a value of type int is pushed from the stack to the local variable position 0, then iInc is executed to add the value of position 0 +1, and then the local variable position 0 is pushed to the output operation
So the value of i1 is 11
I2 is pushed with a value of 10 (bipush), then a value of type int is pushed from the stack to the local variable position 1, then iInc is executed to change the value of position 1 +1, and then the local variable position 1 is pushed to perform the output operation
So the value of i2 is 11
conclusion
Since there is no assignment, there is not much difference.
Problem of the second kind
I3 is first pushed to the location of local variable table 2, and then it is pushed. Execute iinc to increase the value of position 2 by one, and I4 is stored to the location of local variable table 3
So i3 is 11, i4 is still 10
I5 is pushed and stored at the location of local variable table 4. Since it is ++ I, iInc first increments the value of position 4 by one, and then pushes the value of local variable table 4 to perform the assignment operation, so both are 11
The third category of questions
I7 is pushed first, and then stored on local variable table 6. I6 is pushed first, and then the value at 6 is incremented by one. Since this value is stored on local variable table 6, overwrite is generated and the value is changed to 10.
++ I does not create overwrite by performing the increment and then pushing the value onto the stack, in the assignment to the local variable table, so i8 is 11.
The fourth category of questions
I9 =10 is pushed first, and then there is the position of local variable table 8
int i10 = i9++ + ++i9;
Copy the codeIload pushes i9 at 8 and then iInc increses i9 at 8 and then ++i9, increses i9 at 8
At this point, i9=10+1+1 is 12, and then i9 at position 8 is pushed onto the stack. Execute add to add the two i9 in the stack and store the obtained value to the position of local variable table 9
So i10=10+12(i9++ is still 10,++i9 is 12, because I did iinc twice)
The virtual and static methods are then called, and the output statement is executed by pushing the value at 9 onto the stack
Original link: blog.csdn.net/demo_yo/art…
Copyright notice: This article is an original article BY CSDN blogger “Unknown Coder”. It is in accordance with CC 4.0 BY-SA copyright agreement. Please attach the original source link and this statement.
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