I am trying to run the following code in PHP through localhost, but its giving the unexpected output!
<?php
$a = 1;
echo ($a+$a++); // 3
?>
//answer is 3 but answer should be 2 due to post increment here is another code and it gives the same answer! why?
<?php
$a = 1;
echo ($a+$a+$a++);
?>
//answer is still 3 !!!
The PHP manual says the following:
Operator precedence and associativity only determine how expressions are grouped, they do not specify an order of evaluation. PHP does not (in the general case) specify in which order an expression is evaluated and code that assumes a specific order of evaluation should be avoided, because the behavior can change between versions of PHP or depending on the surrounding code.
So what this comes down to, PHP doesn't explicitly define what the end-result is of those types of statements, and it may even change between PHP versions. We call this undefined behavior, and you shouldn't rely on it.
You might be able to find an exact reason somewhere in the source why this order is chosen, but there might not be any logic to it.
Your two examples are being evaluated as follows:
<?php
$a = 1;
echo ($a + $a++); // 3
?>
Really becomes:
<?php
$a = 1;
$b = $a++;
echo ($a + $b); // a = 2, b = 1
?>
Your second example:
<?php
$a = 1;
echo ($a + $a + $a++); // 3
?>
Becomes:
<?php
$a = 1;
$b = $a + $a;
$a++;
echo $b + $a; // 3
?>
I hope this kind of makes sense. You're right that there's no hard logic behind this.
$a+$a++ - 3v4l.org/9Wp8V as if it's evaluated as pre not post. So it makes the OPs example 2 look off. –
Parietal BEHAVIOR OF INCREMENT OPERATION IN SAME LINE WITH CALCULATION IS NOT DEFINIED!
Compilator can generate different code then you expect.
Simple answer from my teacher:
NEVER USE INCREMENT/DECREMENT OPERATOR IN SAME LINE WITH CALCULATIONS!
It's behavior is undefined - computers calculate in different order then humans.
GOOD:
$d = $i++;
$i++;
for ($i = 0; $i < 5; $i++)
CAN BE A PROBLEM (you can read it wrong):
$d = $array[$i++];
WILL BE A PROBLEM:
$d = $i++ + 5 - --$k;
EDIT:
I wrote same code in C++. Checked Assembler code and result is as I said: Math with increment is not logic for people, but you can't say that someone implemented it wrong.
As someone posted in comment:
line #* E I O op fetch ext return operands
-------------------------------------------------------------------------------------
2 0 E > ASSIGN !0, 1
3 1 POST_INC ~2 !0
2 ADD ~3 !0, ~2
3 ECHO ~3
16 4 > RETURN 1
//$a = 1;
//echo ($a+$a++);
line #* E I O op fetch ext return operands
-------------------------------------------------------------------------------------
2 0 E > ASSIGN !0, 1
3 1 ADD ~2 !0, !0
2 POST_INC ~3 !0
3 ADD ~4 ~2, ~3
4 ECHO ~4
5 > RETURN 1
//$a = 1;
//echo ($a+$a+$a++);
Translated to human language:
$a = 1;
echo ($a + $a++);
// After changing to 'computer logic' (ASM):
$c = 1; // for increment operation
$b = $a; // keep value 'before' add +1 operation
$a += $c; // add +1 operation
$d = $a + $b; // calculate value for 'echo'
echo $d; // print result
$d = $array[$i++] is a perfectly common idiom. –
Sanguinary $d = $array[$i++] you shouldn't be a programmer –
Sanguinary $d = $i++ + 5 - --$k; - whilst that might not be particularly readable (and I wouldn't write that myself), it is actually perfectly well defined because none of the variables that are using increment / decrement operators are re-used anywhere else in the expression. If you're being taught otherwise, they're wrong. –
Sanguinary Its because the ++ sign is an incremental operator to a variable. So your
$a = 1
$a++ = 2
($a+$a++) = (1+2) = 3
Thats why it shows 3 as the answer.
© 2022 - 2024 — McMap. All rights reserved.
echo $a;at the end to see the actual effect. I assume it has todo with operator precedence, see Example #2. Which version of PHP are you on? – Sandalwood$a+$a++breaks. eval.in/917220 – Fervor$a = 1; echo $a+ ++$a; //4figure that one out ;p – Parietal$aadded to the first example:$a = 1; echo ($a+$a++); echo "_"; echo $a;. Surprise! It yields3_2. It seems that that behavior occurs from PHP version 5.1. – Nephridium