Why difference in outputs?

[684425]

<?php
$i = 2;
$i = $i++;
echo $i;
?>

Output of above code is 2.

<?php
$i = 2;
$p = $i++;
echo $i;
?>

Output of above code is 3.

 

Why this difference? I mean how these codes are processed to give an output?

[Ch0cu3r]

$i++ increments the value  after the (assignment) operation has occurred, not during the (assignment) operation. For that you need to use  ++$i

 

 But there is no need to do  $i = $i++;  just do  $i++;

[684425]

Actually the question was asked in a facebook group. and my answer there was...

You ignored variable $p in second code and directly called an output of $i so it became $i = 2; $i++; for php.
In your 1st code you are trying to do the same thing, but because of only one variable $i on both sides in 2nd line, php ignored #YOU and did it work. 
1st code
$i = 2; //declared an assigned a value.
$i = $i++; //another declaration.
------------------------------------
2nd code
$i = 2; //declared an assigned a value.
$i++; // direct increment.

Still i was confused... Thanks dear for reply

[Barand]

To make your second code equivalent to the first you would echo $p.

<?php
$i = 2;
$p = $i++;       // value assigned to $p then $i is incremented
echo $p;         //--> 2
echo $i;         //--> 3
?>

[684425]

Explained by a friend.

$i = $i++;    // internally is more like
$temp = $i++; // $temp = 2; $i=3
$i=$temp;     // $i=2

[scootstah]

Explained by a friend.

$i = $i++;    // internally is more like
$temp = $i++; // $temp = 2; $i=3
$i=$temp;     // $i=2

That doesn't seem to be explaining anything.

 

$i = 2;
$p = $i++;

When you say this, you are assigning $p to the value of $i before $i is incremented. So since the value of $i is 2, $p is equal to 2. After $p is assigned to the value of $i, only then does $i get incremented. On the next statement, $i is equal to 3, since it has then been incremented.

 

Now, when you use the same variable, you are basically overriding the would-be incremented value. So if you say

$i = 2;
$i = $i++;

you are setting $i to the value of $i again before it is incremented, which is 2. Since you're using the same variable name, you're effectively preventing the increment from happening.

 

But if you changed it to ++$i, then you would end up with 3.

$i = 2;
$i = ++$i;
echo $i; // $i = 3

This is because, the increment is evaluated before the assignment happens. So by the time $i gets its new value, the increment has happened and the new value is 3.

 

Hopefully that all came out in a way that makes sense.

 

EDIT: This is kind of what is happening internally when you do $i++ vs $++i.

 

// -- $p = $i++;
$i = 2;
$p = $i;
// $i == 2; $p == 2;
$i = $i + 1;
// $i == 3; $p == 2;


// -- $p = $++i;
$i = 2;
$i = $i + 1;
// $i == 3;
$p = $i;
// $i == 3; $p == 3;

Edited January 23, 2015 by scootstah