Weird PHP error: 'Can't use function return value in write context'

Question

I'm getting this error and I can't make head or tail of it.

The exact error message is:

Fatal error: Can't use function return value in write context in /home/curricle/public_html/descarga/index.php on line 48

Line 48 is:

if (isset($_POST('sms_code') == TRUE ) {

What could be going on here?

Here's the full function:

function validate_sms_code() {

    $state = NOTHING_SUBMITED;

    if (isset($_POST('sms_code') == TRUE ) {
        $sms_code = clean_up($_POST('sms_code'));
        $return_code = get_sepomo_code($sms_code);

        switch($return_code) {

          case 1:
            //no error
            $state = CORRECT_CODE;
            break;

          case 2:
            // code already used
            $state = CODE_ALREADY_USED;
            break;

          case 3:
            // wrong code
            $state = WRONG_CODE;
            break;

          case 4:
            // generic error
            $state = UNKNOWN_SEPOMO_CODE;
            break;

          default:
            // unknown error
            $state = UNKNOWN_SEPOMO_CODE;
            throw new Exception('Unknown sepomo code: ' . $return_code);
            break;
        }

    } else {
        $state = NOTHING_SUBMITED;
    }
    dispatch_on_state($state);
}

$_POST('sms_code') should be $_POST['sms_code'], by the way. — davidtbernal, Commented Oct 7, 2009 at 16:25
Also, the else clause on the if statement is not needed. $state will not be modified, no need to reset it to NOTHING_SUBMITED if it's already that value. — Glenn Dayton, Commented Aug 15, 2012 at 17:13
People of the future, can anyone explain what the message actually means? — OGHaza, Commented Jun 22, 2017 at 0:02

Fabian · Accepted Answer · 2016-04-19 13:18:18Z

497

This also happens when using empty on a function return:

!empty(trim($someText)) and doSomething()

because empty is not a function but a language construct (not sure), and it only takes variables:

Right:

empty($someVar)

Wrong:

empty(someFunc())

Since PHP 5.5, it supports more than variables. But if you need it before 5.5, use trim($name) == false. From empty documentation.

edited Apr 19, 2016 at 13:18

Fabian

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answered Jan 31, 2010 at 21:04

rolfen

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Dang, I just got this error too, doing your exact example (trim inside empty). Thanks x3. What a strange error.. I still don't fully understand :/ empty() can take a string.. and trim() returns a string.. so wth?
– Nick Rolando
Commented Feb 8, 2012 at 17:23
15

@Shredder - It's because empty() isn't actually a function; it's a language construct, like echo. The PHP parser handles language constructs differently. In the case of empty, it looks at the parameter as a variable, not something to be evaluated, so if you try to pass a function, it breaks. At least that's my understanding of it. More info here and here.
– grant
Commented Jun 26, 2012 at 21:04
1

This is the exact error I ran into as well. Why isn't this the top answer!?
– Adam Fowler
Commented Feb 22, 2013 at 17:04
2

My colleague was getting this error while I was not. Confusion ensued, until I checked the docs - as of PHP 5.5, empty() will accept the return from a function as well as a variable. us3.php.net/empty
– James Alday
Commented Feb 26, 2014 at 19:30
1

Horray for 5.5! Until you upgrade, you could throw something like this into your main function include file: function mTEE($val){ return empty($val); }
– TecBrat
Commented Mar 14, 2014 at 15:18

| Show 6 more comments

chaos · Accepted Answer · 2020-02-21 22:25:47Z

113

You mean

if (isset($_POST['sms_code']) == TRUE ) {

though incidentally you really mean

if (isset($_POST['sms_code'])) {

edited Feb 21, 2020 at 22:25

answered Oct 7, 2009 at 16:23

chaos

123k33 gold badges305 silver badges310 bronze badges

3

...or if (isset($_POST['sms_code']) === TRUE ) { :-)
– keyboardSmasher
Commented Apr 14, 2014 at 14:24
I also found the solution for the issue I was facing, so FYI - this goes also for $_REQUEST (parameters in the URL of the page).
– TheCuBeMan
Commented Dec 31, 2015 at 9:37
6

I realize I'm super late to this, but it's also PHP version dependent, right? I'm pretty sure that's legal in later PHP versions like 5.6, but I think it doesn't in 5.3
– UnsettlingTrend
Commented Dec 8, 2016 at 16:14

Add a comment |

GAMITG · Accepted Answer · 2016-01-18 06:42:06Z

22

if (isset($_POST('sms_code') == TRUE ) {

change this line to

if (isset($_POST['sms_code']) == TRUE ) {

You are using parentheseis () for $_POST but you wanted square brackets []

:)

OR

if (isset($_POST['sms_code']) && $_POST['sms_code']) { 
//this lets in this block only if $_POST['sms_code'] has some value

edited Jan 18, 2016 at 6:42

GAMITG

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answered Oct 7, 2009 at 16:24

TigerTiger

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1

Nope, you cannot write "if (isset($_POST['sms_code'] == TRUE ) {", there's a missing ")".
– middus
Commented Oct 7, 2009 at 16:26
1

+several billion cool points for the phrae 'you are using parenthesis...but you wanted square brackets', which was what my problem (that led me to this question) was
– Kevin Horn
Commented Mar 22, 2012 at 14:36
1

Phew, thx man! Now, if the parser would say "I expected square brackets after an array variable's name, duh!", it would sound much cooler than "Can't use function return value in write context." I might submit that to the PHP guys as a suggestion.
– Joe Völker
Commented Dec 25, 2012 at 9:40

Add a comment |

T.Todua · Accepted Answer · 2017-01-08 17:49:38Z

13

for WORDPRESS:

instead of:

if (empty(get_option('smth')))

should be:

if (!get_option('smth'))

edited Jan 8, 2017 at 17:49

answered Mar 26, 2015 at 9:59

T.Todua

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T.Todua · Accepted Answer · 2017-11-18 08:15:11Z

11

Correct syntax (you had a missing parentheses in the end):

if (isset($_POST['sms_code']) == TRUE ) {
                            ^

p.s. you dont need == TRUE part, because BOOLEAN (true/false) is returned already.

edited Nov 18, 2017 at 8:15

T.Todua

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answered Oct 7, 2009 at 16:25

middus

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Mehdi Karamosly · Accepted Answer · 2014-10-20 16:34:51Z

5

This can happen in more than one scenario, below is a list of well known scenarios :

// calling empty on a function 
empty(myFunction($myVariable)); // the return value of myFunction should be saved into a variable
// then you can use empty on your variable

// using parenthesis to access an element of an array, parenthesis are used to call a function

if (isset($_POST('sms_code') == TRUE ) { ...
// that should be if(isset($_POST['sms_code']) == TRUE)

This also could be triggered when we try to increment the result of a function like below:

$myCounter = '356';

$myCounter = intVal($myCounter)++; // we try to increment the result of the intVal...
// like the first case, the ++ needs to be called on a variable, a variable should hold the the return of the function then we can call ++ operator on it.

edited Oct 20, 2014 at 16:34

answered May 5, 2014 at 17:28

Mehdi Karamosly

5,3982 gold badges32 silver badges50 bronze badges

Do you want to get bugs? Because this is how you get bugs. No but seriously, don't ever do this. Its like doing if(i++), it may look shorter than incrementing the counter and then evaluating its value, but it is going to cause you headaches if it ever bugs out. Edit: Im not implying that you do it, just pointing it out to new programmers reading the answer and thinking this is a neat thing, to increment values while calculating or evaluating that which is being incremented.
– Victor D.
Commented Oct 1, 2014 at 20:56
@VictorD. I am just saying that error could happen in that situation that means it is not good to do it.
– Mehdi Karamosly
Commented Oct 2, 2014 at 23:56
Can I suggest one more? You do = instead of == in an if statement.
– Josiah
Commented Mar 9, 2016 at 18:19
For me it was PHP version, upgrade to 7.0 :)
– nodws
Commented Apr 2, 2018 at 16:24
still happening on PHP 8.1 intval($strCounter)++
– jesusduarte
Commented Feb 12, 2022 at 3:11

Add a comment |

Warren Sergent · Accepted Answer · 2015-06-09 12:41:07Z

3

The problem is in the () you have to go []

if (isset($_POST('sms_code') == TRUE)

by

if (isset($_POST['sms_code'] == TRUE)

edited Jun 9, 2015 at 12:41

Warren Sergent

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answered May 24, 2015 at 12:01

Diego

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husnixs · Accepted Answer · 2015-08-19 16:08:22Z

3

I also had a similar problem like yours. The problem is that you are using an old php version. I have upgraded to PHP 5.6 and the problem no longer exist.

edited Aug 19, 2015 at 16:08

answered Aug 19, 2015 at 10:41

husnixs

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tomelin5 · Accepted Answer · 2017-06-23 04:13:40Z

1

Another scenario where this error is trigered due syntax error:

ucwords($variable) = $string;

answered Jun 23, 2017 at 4:13

tomelin5

551 silver badge7 bronze badges

1

Is because ucwords return a string and the context of calling is incorrect, if you try for example with $Test = ''; ${ucwords('test')} = 'String new !'; echo $Test;, then show you the new value assigned.
– kip
Commented Sep 16, 2017 at 18:17

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Shaakir · Accepted Answer · 2016-07-30 08:23:13Z

0

i also ran into this problem due to syntax error. Using "(" instead of "[" in array index:

   foreach($arr_parameters as $arr_key=>$arr_value) {
        $arr_named_parameters(":$arr_key") = $arr_value;
    }

answered Jul 30, 2016 at 8:23

Shaakir

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user12845085 · Accepted Answer · 2020-02-17 13:47:28Z

This error is quite right and highlights a contextual syntax issue. Can be reproduced by performing any kind "non-assignable" syntax. For instance:

function Syntax($hello) { .... then attempt to call the function as though a property and assign a value.... $this->Syntax('Hello') = 'World';

The above error will be thrown because syntactially the statement is wrong. The right assignment of 'World' cannot be written in the context you have used (i.e. syntactically incorrect for this context). 'Cannot use function return value' or it could read 'Cannot assign the right-hand value to the function because its read-only'

The specific error in the OPs code is as highlighted, using brackets instead of square brackets.

TheTechGuy · Accepted Answer · 2020-04-04 18:00:42Z

-2

Can be cause by wrong operator, =, when it should be ==

if(mysql_num_rows($result) = 1)
    return $result;
else
    return false;

This code throws this error

Note that = is assignment operator and not comparison operator. Fix is to change = to ==.

answered Apr 4, 2020 at 18:00

TheTechGuy

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