My understanding of C # expressions

This morning I saw "do you really know a ^ = (B ^ = (a ^ = B)?" Article,ArticleAn interesting expression: A ^ = (B ^ = (a ^ = B );Program: 

1 Int A =   2 ;
2 Int B =   9 ;
3 A ^ = (B ^ = ( ^ = B ));
4 Console. writeline ( " {0}, {1} " , A. tostring (), B. tostring ());

The final calculation result is not expected 9, 2, but 0, 2. There is also one expression: 

1 Int I = 0 ;
2 Int J = (I ++ ) + (I ++ );
3 J =?

What is the J value? 2? No. The correct result is 1.

Why is there such a result? Is it a compiler bug? In fact, the compiler may not be a mathematical idiot, but our understanding of the computer is incorrect.

In C #, expression calculation follows a rule:Calculated from left to right.

For example, Z = x operation Y. The computer calculates the values of x and y in sequence, then operationx and Y, and finally assigns the values to Z.

Similarly, if z = x operation1 (x operation2 y), the computer regards (x operation2 y) as a variable temp. In this case, Z = x operation1 temp. Then, the computer calculates the temp expression, presses the stack, computes the temp expression, and outputs the result. This is a recursive process. Note: In this process, X and temp are two parallel expressions. If X is reoperated in temp, the result of X is not affected. Therefore, the expression rule is calculated from left to right, and X is left. The calculation is complete. If it is Z = temp operation1 X, the operation on X in temp will affect X itself.

Okay. After theoretical analysis, let's take a look at why the results of the above two expressions are surprising.

For the first expression: A ^ = (B ^ = (a ^ = B )). First, the operator is decomposed into a = a ^ (B = B ^ (A = a ^ B), which makes it clearer.

1. The computer decomposition expression is a = a ^ temp1. Calculate a on the left. A is a variable, so a is replaced by 2. Next, we calculated the temp1 expression on the right and found that it was complicated and could not be directly replaced by a value. Therefore, we started to calculate the temp1 stack.

2. temp1 is B = B ^ (A = a ^ B ). On the left side of the calculation, B is a variable and can be directly assigned to 9. The expression on the right is replaced by temp2.

3. temp2 is a = a ^ B. On the left side of the calculation, A is a variable, and 2 is replaced; B is also a variable, and 9 is replaced. And calculate temp2 = 2 ^ 9 = 11.

4. regression calculation Temp 1 = 9 ^ temp2 = 9 ^ 11 = 2, that is, B = 2.

5. regression calculation A = 2 ^ temp1 = 2 ^ 2 = 0. That is, a = 0.

OK. The result of 0 and 2 is displayed.

For the second expression: J = (I ++) + (I ++ );.

1. Calculate the expression on the left, I ++. Because it is a suffix, the (I ++) expression returns I. At this time, I = 0, meaning 0 is returned.

2. Execute the ++ operator. 0 ++ is 1, So I = 1;

3. Calculate the expression (I ++) On the right, which is also a suffix. Therefore, I is returned. At this time, I = 1, which means 1 is returned;

4. Execute the ++ operator. 1 ++ is 2, so I = 2;

5 is assigned to J. J = 0 + 1 = 1.

In summary, it is not a compiler bug, but a computer's practice will always be a dead-end.

And,In actual projects, do not use such expressions. Instead, write a few more lines and do not pile them in one line. We must follow 《CodeSpecifications.This is also to improve the readability and maintainability of the Code.

Finally, I will give you a question:

Int I = 0; Int J = (I --) + (I ++) + (++ I); what are the calculation results of I and J?