In depth explanation of i = i + + and i = + + i in java

public class Count {
 public static void main(String[] args) {
 int i = 0;
 i = i++ ;
 System.out.println(i);
 }
}

The i output from the above code is 0. If i = i + + is replaced by i = + + i, it will output 1 again. This is because i + + is assigned first and then calculated, but why assign first?

public static void main(String[] args) {
 int i = 0;
 i++ ;
 }

 public static void main(java.lang.String[]);
 Code:
 0: iconst_0  Will int Type 0 Push to the top of the stack (find 0 ) 
 1: istore_1  Put the top of the stack int Type value is stored in the first 2 Local variables (variables are i ) 
 2: iinc   Will specify the int Type variable increases the specified value ( Such as i++, i--, i+=2 Etc )
 5: return

The above figure shows the bytecode of i + +, 1 touch 1 sample of + + i and i + +, 0 and 1 are the most basic values for i, and iinc is to increase i by itself.

But what if you add 1 i = i + +?

// i = i++;
 public static void main(java.lang.String[]);
 Code:
 0: iconst_0  Will int Type 0 Push to the top of the stack 
 1: istore_1  Put the top of the stack int Type value is stored in the first 2 Local variables 
 2: iload_1  Will be 2 A int Type local variables are pushed to the top of the stack 
 3: iinc   Will specify the int Type variable increases the specified value ( Such as i++, i--, i+=2 Etc )
 6: istore_1  Put the top of the stack int Type value is stored in the first 2 Local variables 
 7: return

0 and 1 are still assigned to i as 0, 2 is to push the current value of i to the top of the stack, then iinc increases i of the local variable table, and 6 is to assign the value of the top of the stack to i. Note that when 2 is pushed to the top of the stack, it is 0, so it changes back to 0.

// i = ++i
 public static void main(java.lang.String[]);
 Code:
 0: iconst_0  Will int Type 0 Push to the top of the stack 
 1: istore_1  Put the top of the stack int Type value is stored in the first 2 Local variables 
 2: iinc   Will specify the int Type variable increases the specified value ( Such as i++, i--, i+=2 Etc )
 5: iload_1  Will be 2 A int Type local variables are pushed to the top of the stack 
 6: istore_1  Put the top of the stack int Type value is stored in the first 2 Local variables 
 7: return

When i = + + i, iinc goes one step ahead, indicating that before the value of i is pressed to the top of the stack, it is self-increasing, so the value of the top of the stack is 1. At this time, it is self-increasing when it is assigned to i in the local variable table.

Summary: The iinc operation is directed at the local variable table, Without going through the operation stack, the so-called pre-assignment of i + + means that i stores the original value into the operand stack before self-increment from the bytecode level. Therefore, if b = i + + is executed, the reason why b is equal to the original value of i is because the value of the operand stack is the value before self-increment, while the self-increment operation of + + i is advanced, so after b = + + i, the value of b and i are 1.

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