PHP Strange Parsing method

Code One:
$i = 1;
$i = $i + + + + $i;
echo $i;

Code two:
$i = 1;
$i = + + $i + $i + +;
echo $i;

Both of these codes result in 4.

I don't think I can understand the parsing principle of PHP to ask Big Brother pointing

Reply to discussion (solution)

What's wrong with it?
$i = 1;
$i = $i + + + + $i;
==>
$i = 1 (but because of $i++, so $i is 2) + 3 (+ + $i $i is 2, plus 1 is 3)

$i = 1;
$i = + + $i + $i + +;
==>
$i = 2 + 2

What's wrong with it?
$i = 1;
$i = $i + + + + $i;
==>
$i = 1 (but because of $i++, so $i is 2) + 3 (+ + $i $i is 2, plus 1 is 3)

$i = 1;
$i = + + $i + $i + +;
==>
$i = 2 + 2

Do the + + $i first
And the addition should be a.
Last $i + +
It's supposed to be 5.

$i = 1;
$i = $i + +;
echo $i;
echo $i;

Output two X 1

+ + operation was swallowed.

The reason for the seemingly returning 4 is that the last $i + + is swallowed up at $i = ...

The reason for the seemingly returning 4 is that the last $i + + is swallowed up at $i = ...
Your understanding is not the same as mine.
That's how I analyze it.
$i = $i + +
First you have to admit that this is an assignment operation that assigns the right value to the left (that is, $i)

First look to the right, the current value of I is 1, and the value of i++ itself is 1, but the value of I after i++ operation is 2.
The point is finally, to replicate. It says that i++ itself is 1, copy 1 to I, and I will change from 2 to 1.
Did not swallow a say.

Right! You understand the right thing!

$i = 1; $ii = + + $i + $i ++;echo "\ $i = $i \ $ii = $ii";//$i =3 $ii =4$i = 1; $ii = $i ++;echo ' $i = '. $i; $i =2echo ' $ii = '. $ii; $ii =1

The reason for the seemingly returning 4 is that the last $i + + is swallowed up at $i = ...
Your understanding is not the same as mine.
That's how I analyze it.
$i = $i + +
First you have to admit that this is an assignment operation that assigns the right value to the left (that is, $i)

First look to the right, the current value of I is 1, and the value of i++ itself is 1, but the value of I after i++ operation is 2.
The point is finally, to replicate. It says that i++ itself is 1, copy 1 to I, and I will change from 2 to 1.
Did not swallow a say.


First, assign the value after + +

$i = $i + +;

Seems to have been interpreted like this.
$i = Funcation () {return $i + +;}

The + + operation is treated as a local variable

The foundation did not play well. Study the difference between pre-plus and post-plus.

In this case, the + + as a function, ++a return is a+1, and a++ return is a. However, after two operations, A is a+1, knowing that the return value is not different.

In this case, the + + as a function, ++a return is a+1, and a++ return is a. However, after two operations, A is a+1, knowing that the return value is not different.

Both in $i + +

The result should be $i =2

$j = $i + +;
$i = $i + +;

Not the same for the result $i

$i = $i + + This is undefined behavior (Undefined behavior) See Http://stackoverflow.com/questions/1998903/php-i-i-crashed-the-server

The reason for the seemingly returning 4 is that the last $i + + is swallowed up at $i = ...
No, you have to understand that one is performed after the operation, one is the operation after the execution, you understand the wrong $i++ and + + $i Difference

$i + + is the first to take plus, + + $i, this is the first plus the value
$i = $i + + + + $i;
This is supposed to be
$i = 1 (first fetch and then add, so here is still 1) + 3 (because before $i++ got 2, here first add again, so it is 3);

The first one is the
The second one is 1+3.
So it's all equal to 4.
$i + + This expression is to be $i +1, and this expression value is also $i+1,
+ + $i This expression will also $i +1, but the value of this expression is $i, not $i+1, understand?

This is a two different expression.

The first one is the
The second one is 1+3.
So it's all equal to 4.
$i + + This expression is to be $i +1, and this expression value is also $i+1,
+ + $i This expression will also $i +1, but the value of this expression is $i, not $i+1, understand?

This is a two different expression.

Dizzy, said the reverse ...

The first one is 1+3.
And the second one is the
So it's all equal to 4.
+ + $i This expression is to be $i +1, and this expression value is also $i+1,
$i + + This expression will also $i +1, but the value of this expression is $i, not $i+1, understand?

This is a two different expression.

Did not learn how long, personally think it should be so understood, $i + + + + + First 1 to add + + $i (2) finally equals 3, because the front of the $i++ has not been self-increasing, the entire expression after the run, $i there is a self-increment, so is 4

Maybe that's what you got.
$i = 1;
$j = $i + + + + $i;
echo $i. ': ' $j;

$i = 1;
$j = + + $i + $i + +;
echo $i. ': ' $j;

Seems to be because

$i = $i + +;

Be resolved to

$i = $i; $i + 1;

Instead of

$i = $i; $i = $i +1;


Maybe both of them have a 1+3.

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