+ + And a little doubt-php Tutorial

+ + And a little doubt $ A = 1;
$ B = & $;
$ B = $ a ++;

// Calculate $ a and $ B
In the second sentence, B also points to the address of;
In the third sentence, the value of a is added with 1, so $ B = 1;
The key is that I don't know $ a ++ later. I don't mean to add 1 after using it, so now $ a = 2;
But the address of $ B is the same as the address of a. isn't the address of $ B equal to 2? Or is $ a ++ not executed in the third sentence? The answer to this question is $ a and $ B is 1;
For example
$ A = 1;
$ B = $ a ++;
?>
This is not $ B = 1, $ a = 2
A little dizzy...

Reply to discussion (solution)

$ A = 1;
$ B = $ a ++;

Result:
$ B = 1
$ A = 2;

If $ B = & $;
Indicates that $ a $ B points to the same address.
If $ B changes, the value of $ a also changes, and eventually $ B = 1, $ a = 1

$a = 1;$b = $a++;var_dump($a, $b);/*int(2)int(1)*/$a = 1;$b = &$a;$a++;var_dump($a, $b);/*int(2)int(2)*/$a = 1;$b = &$a;$b = $a++;var_dump($a, $b);/*int(1)int(1)*/

For $ B = $ a ++;
Assign the value 1 of $ a to $ B, but the priority of ++ is higher than =, so the value is not assigned first, but ++, therefore, the value of $ a is 2 (because $ B is a reference of $ a, the value of $ B is also 2)
Then assign a value. The value of $ B is equal to 1.
Because $ B is a reference of $ a, $ a is changed back to 1.

The basics are not well laid ~ Thank you!

If you do not know the operator priority, you can understand it!

   
The image point is that a B represents several apples in the same drawer! James is in charge of taking a look at the drawer where a is located and remembering a few apples in the drawer (that is, what is a), and then placing an apple in the drawer! Then, change the number of apples in the drawer where B is located to the number of just remembered apples, that is, the value of! So change it! It's changed back to!
 
The first case is not referenced.
  
$ A = 1; $ B = $ a; // B = 1 a = 1 $ B = $ a ++; // a is assigned to B first, then auto-increment B = 1 a = 2 echo $ a; echo $ B;
  
  
In the second case, reference
  
$ A = 1; $ B = & $ a; // B references a, that is, a, and B points to the same memory address $ B = $ a ++; // a reads the value and executes auto-increment. a = 2 B = 2, and then assigns the value to B. Therefore, B = 1 a = 1 echo $; echo $ B;