php - What impact will the ampersand reference pass have on the operation results?

Question

$a = 1;
$b = &$a; //Pass address. Commenting out this line returns 5. If it is not commented out (PHP version <7), it returns 6
$c = ( $a) ( $a);
echo $c;
Does it have anything to do with the version? ? Also, the result is 6, which is really puzzling to me

Answer 1

Let’s first talk about the operation of adding reference assignment

$b = &$a;

For this operation, it can be considered that both $a and $b point to the memory where the original $a variable is located (for convenience of explanation later, it is called memory X), that is to say, any subsequent The operations of $a or $b are directly modifying the value in this memory.

So the running process after adding this line:

1. The previous steps are omitted, the initial a is 1

2. $c = (++$a) + (++$a) The two times ++$a are both operating memory twice, so when calculating the value of $c, the value of this memory is taken, that is, the value in memory X after it has been incremented twice. Therefore, it is 3 + 3 = 6 (when calculating $c, the value stored in memory X is 3).

   Let’s talk about the operation after commenting the reference

If no reference is added, then the auto-increment operation of

$a

directly modifies

memory X, and then returns a copy of memory It is not obtained directly from Memory X, but a copy after changing the value of Memory

Answer 2

This is related to the underlying implementation of PHP. It’s a long story. It is recommended to look at the analysis of the same problem on github.
An exploration caused by a PHP bug:
https://github.com/xurenlu/ph...

Answer 3

You output C, what does C have to do with your B?