Let's preface this with an example of a script with typo:
$bc = getBaseConcept();
$bs['key'] = doOtherStuff($bc['key']);
return $bc;
Obviously in the middle line is a typo. It should be $bc instead of $bs. (And yes this was a legit typo of mine just minutes ago before writing this question)
This did not produce a warning.
So my question is: Is there a configuration option that lets this produce a warning?
Specifically: Writing to an array key of a name that was previously undefined.
E_ALL does not seem to help. This should not only generate the warnings for $bar but I want also a warning for $foo.
<?php
ini_set('error_reporting', E_ALL);
echo ini_get('error_reporting'), "\n";
$foo['bar'] = $bar['foo'];
32767
PHP Warning: Undefined variable $bar in Standard input code on line 4
PHP Warning: Trying to access array offset on value of type null in Standard input code on line 4
Unfortunately, php is not like other programming languages. If an unknown variable is used, PHP does the initialisation without complaining. Unfortunately, there is nothing you can do about this.
If you are using an IDE like phpstorm or netbeans the IDE will usually show a hint for uninitisalized variables.
It is not necessary to initialize variables in PHP however it is a
very good practice. Uninitialized variables have a default value of
their type depending on the context in which they are used - booleans
default to false, integers and floats default to zero, strings (e.g.
used in echo) are set as an empty string and arrays become to an empty
array.
https://www.php.net/manual/en/language.variables.basics.php
Related
Edit 2022: This appears to be fixed as of PHP 7.4 which emits a notice.
In PHP, I have error_reporting set to report everything including notices.
Why does the following not throw any notices, errors or anything else?
$myarray = null;
$myvalue = $myarray['banana'];
Troubleshooting steps:
$myarray = array();
$myvalue = $myarray['banana'];
// throws a notice, as expected ✔
$myarray = (array)null;
$myvalue = $myarray['banana'];
// throws a notice, as expected ✔
$myarray = null;
$myvalue = $myarray['banana'];
// no notice or warning thrown, $myvalue is now NULL. ✘ Why?
It's possible it's a bug in PHP, or I'm just not understanding something about how this works.
There are three types which it might be valid to use the array derefence syntax on:
Arrays
Strings (to access the character at the given position)
Object (objects implementing the ArrayAccess interface)
For all other types, PHP just returns the undefined variable.
Array dereference is handled by the FETCH_DIM_R opcode, which uses zend_fetch_dimension_address_read() to fetch the element.
As you can see, there is a special case for NULLs, and a default case, both returning the undefined variable.
Usually, when you try to use a value of one type as if it were another type, either an error or warning gets thrown or "type juggling" takes place. For example, if you try to concatenate two numbers with ., they'll both get coerced to strings and concatenated.
However, as explained on the manual page about type juggling, this isn't the case when treating a non-array like an array:
The behaviour of an automatic conversion to array is currently undefined.
In practice, the behaviour that happens when this "undefined behaviour" is triggered by dereferencing a non-array is that null gets returned, as you've observed. This doesn't just affect nulls - you'll also get null if you try to dereference a number or a resource.
There is an active bug report started at 2006.
And in documentation it is a notice about this in String section.
As of PHP 7.4, this behavior how emits a Notice.
"Trying to access array offset on value of type null"
See the first item in this 7.4 migration page.
https://www.php.net/manual/en/migration74.incompatible.php
This recently struck one of my colleagues in the butt because he neglected to validate the result of a database query before attempting to access column data from the variable.
$results = $this->dbQuery(...)
if($results['columnName'] == 1)
{
// WHEN $results is null, this Notice will be emitted.
}
And I just noticed #Glen's comment, above, citing the relevant RFC.
https://wiki.php.net/rfc/notice-for-non-valid-array-container
Edit 2022: This appears to be fixed as of PHP 7.4 which emits a notice.
In PHP, I have error_reporting set to report everything including notices.
Why does the following not throw any notices, errors or anything else?
$myarray = null;
$myvalue = $myarray['banana'];
Troubleshooting steps:
$myarray = array();
$myvalue = $myarray['banana'];
// throws a notice, as expected ✔
$myarray = (array)null;
$myvalue = $myarray['banana'];
// throws a notice, as expected ✔
$myarray = null;
$myvalue = $myarray['banana'];
// no notice or warning thrown, $myvalue is now NULL. ✘ Why?
It's possible it's a bug in PHP, or I'm just not understanding something about how this works.
There are three types which it might be valid to use the array derefence syntax on:
Arrays
Strings (to access the character at the given position)
Object (objects implementing the ArrayAccess interface)
For all other types, PHP just returns the undefined variable.
Array dereference is handled by the FETCH_DIM_R opcode, which uses zend_fetch_dimension_address_read() to fetch the element.
As you can see, there is a special case for NULLs, and a default case, both returning the undefined variable.
Usually, when you try to use a value of one type as if it were another type, either an error or warning gets thrown or "type juggling" takes place. For example, if you try to concatenate two numbers with ., they'll both get coerced to strings and concatenated.
However, as explained on the manual page about type juggling, this isn't the case when treating a non-array like an array:
The behaviour of an automatic conversion to array is currently undefined.
In practice, the behaviour that happens when this "undefined behaviour" is triggered by dereferencing a non-array is that null gets returned, as you've observed. This doesn't just affect nulls - you'll also get null if you try to dereference a number or a resource.
There is an active bug report started at 2006.
And in documentation it is a notice about this in String section.
As of PHP 7.4, this behavior how emits a Notice.
"Trying to access array offset on value of type null"
See the first item in this 7.4 migration page.
https://www.php.net/manual/en/migration74.incompatible.php
This recently struck one of my colleagues in the butt because he neglected to validate the result of a database query before attempting to access column data from the variable.
$results = $this->dbQuery(...)
if($results['columnName'] == 1)
{
// WHEN $results is null, this Notice will be emitted.
}
And I just noticed #Glen's comment, above, citing the relevant RFC.
https://wiki.php.net/rfc/notice-for-non-valid-array-container
I want to use a global variable setup where they are all declared, initialized and use friendly syntax in PHP so I came up with this idea:
<?
error_reporting(E_ERROR | E_WARNING | E_PARSE | E_NOTICE);
$GLOBALS['debugger'] = 1; // set $GLOBALS['debugger'] to 1
DEFINE('DEBUGGER','$GLOBALS["debugger"]'); // friendly access to it globally
echo "1:" . DEBUGGER . ":<br>";
echo "2:" . ${DEBUGGER}. ":<br>";
echo "3:" . $GLOBALS['debugger'] . ":<br>";
if (DEBUGGER==1) {echo "DEBUG SET";}
?>
generates the following:
1:$GLOBALS["debugger"]:
Notice: Undefined variable: $GLOBALS["debugger"] in /home/tra50118/public_html/php/test.php on line 8
2::
3:1:
How can there be an error with 2: when clearly $GLOBALS["debugger"] IS defined? And then not generate a similar notice with the test at line 10?
I think what I am trying to do is to force PHP to interpret a string ($GLOBALS["debugger"]) as a variable at run time i.e. a constant variable variable
Disclaimer: I agree with the comments, globals are generally a bad idea.
That said, there's a few questions here that are worth answering, and the concept of indirection is useful, so here goes.
${'$GLOBALS["debugger"]'} is undefined. You don't include the leading '$' when using indirection. So, the correct version would be define('DEBUGGER', 'GLOBALS["debugger"]').
But, this doesn't work either. You can only access one level down via indirection. So you can access the array $GLOBALS, but you can't access keys in that array. Hence, you might use :
define('DEBUGGER', 'debugger');
${DEBUGGER};
This isn't useful, practically. You may as well just use $debugger directly, as it's been defined as a global and will be available everywhere. You may need to define global $debugger; at the start of functions however.
The reason your if statement is not causing notices is because you defined DEBUGGER to be a string. Since you aren't trying to use indirection in that line at all, it ends up reading as:
if ("$GLOBALS['debugger']"==1) {echo "DEBUG SET";}
This is clearly never true, though it is entirely valid PHP code.
I think you may have your constants crossed a bit.
DEFINE('DEBUGGER','$GLOBALS["debugger"]'); sets the constant DEBUGGER to the string $GLOBALS["debugger"].
Note that this is neither the value nor the reference, just a string.
Which causes these results:
1: Output the string $GLOBALS["debugger"]
2: Output the value of the variable named $GLOBALS["debugger"]. Note that this is the variable named "$GLOBALS["debugger"]", not the value of the key "debugger" in the array $GLOBALS. Thus a warning occurs, since that variable is undefined.
3: Output the actual value of $GLOBALS["debugger"]
Hopefully that all makes sense.
OK, thanks to all who answered. I think I get it now, I am new to PHP having come form a C++ background and was treating the define like the C++ #define and assuming it just did a string replace in the precompile/run phase.
In precis, I just wanted to use something like
DEBUGGER = 1;
instead of
$GLOBALS['debugger'] = 1;
for a whole lot of legitimate reasons; not the least of which is preventing simple typos stuffing you up. Alas, it appears this is not doable in PHP.
Thanks for the help, appreciated.
You can not use "variable variables" with any of the superglobal arrays, of which $GLOBALS is one, if you intend to do so inside an array or method. To get the behavior you would have to use $$, but this will not work as I mentioned.
Constants in php are already global, so I don't know what this would buy you from your example, or what you are going for.
Your last comparison "works" because you are setting the constant to a string, and it is possible with PHP's typecasting to compare a string to an integer. Of course it evaluates to false, which might be surprising to you, since you expected it to actually work.
Stupid question - I'm surprised this one has bitten me. Why do undefined constants in PHP evaluate to true?
Test case:
<?php
if(WHATEVER_THIS_ISNT_DEFINED)
echo 'Huh?';
?>
The above example prints 'Huh?'
Thanks so much for your help! :)
Try defined('WHATEVER_THIS_ISNT_DEFINED')
When PHP encounters a constant that is not defined, it throws an E_NOTICE, and uses the constant name you've tried to use as a string. That's why your snippet prints Huh!, because a non-empty string (which is not "0") will evaluate to true.
From the manual:
If you use an undefined constant, PHP
assumes that you mean the name of the
constant itself, just as if you called
it as a string (CONSTANT vs
"CONSTANT"). An error of level
E_NOTICE will be issued when this
happens.
If you set your error reporting level to report E_NOTICEs, which is a good practice during development, you will also see the notice thrown.
PHP Constant Syntax
defined()
Casting to Boolean
error_reporting
error_reporting() function
From the manual:
If you use an undefined constant, PHP assumes that you mean the name of the constant itself, just as if you called it as a string (CONSTANT vs "CONSTANT").
Basically, if WHATEVER_THIS_ISNT_DEFINED isn't defined, PHP interprets it as "WHATEVER_THIS_ISNT_DEFINED". Non-empty strings evaluate to true, so your expression will always pass (unless WHATEVER_THIS_ISNT_DEFINED is defined and set to a falsey value.)
This is, frankly, stupid behaviour. It was implemented, I believe, to allow things like $foo[bar] to work when the programmer should have used $foo['bar']. It's illogical behaviour like this that makes people think PHP isn't a real programming language.
The way to test whether a constant is defined is with defined.
Undefined constants are treated as strings by PHP: docs. Taking that fact, think it through in English language:
If "WHATEVER_THIS_ISNT_DEFINED", then do something.
... it is logical that it is "true" - you aren't comparing anything to anything else.
That is why, when doing if statements, it is best practice to include a specific evaluation. If you're checking for false, put it in the code: if (something === false) vs if (something). If you're checking to see if it is set, use isset, and so on.
Also, this highlights the importance of developing with notices and warnings enabled. Your server will throw a notice for this issue:
Notice: Use of undefined constant
MY_CONST - assumed 'MY_CONST' in
some_script.php on line 5
Turn on notices and warnings to develop, turn them off for production. Can only help!
Try defined(). If it's not defined then the constant assumes it's simply text.
Note that constant name must always be quoted when defined.
e.g.
define('MY_CONST','blah') - correct
define(MY_CONST,'blah') - incorrect
also
<?php
if (DEBUG) {
// echo some sensitive data.
}
?>
and saw this warning:
"Use of undefined constant DEBUG - assumed 'DEBUG'"
A clearer workaround is to use
<?php
if (defined('DEBUG')) {
// echo some sensitive data.
}
?>
See http://php.net/manual/en/language.constants.php
It's not just constants, it is a much broader issue with PHP's parsing engine. (You ought to see warnings in your logs.)
In PHP, "bare words" that it doesn't recognize are generally treated as strings that happen to be missing their quotes, and strings with a non-zero length tend to evaluate to true.
Try this:
$x = thisisatest ;
$y = "thisisatest";
if($x == $y){
echo("They are the same");
}
You should see "They are the same".
Old question, but in addition to defined() you can also use strict type checking using ===
<?php
if(WHATEVER_THIS_ISNT_DEFINED === true) // Or whatever type/value you are trying to check
echo 'Huh?';
I haven't made any changes to the code affecting the COOKIE's and now I get the following:
Use of undefined constant COOKIE_LOGIN - assumed 'COOKIE_LOGIN'
//Destroy Cookie
if (isset($_COOKIE[COOKIE_LOGIN]) && !empty($_COOKIE[COOKIE_LOGIN]))
setcookie(COOKIE_LOGIN,$objUserSerialized,time() - 86400 );
I'm not sure what I need to do to actually change this since I do not know what chnaged to begin with and so cannot track the problem.
Thanks.
You need to surround the array key by quotes:
if (isset($_COOKIE['COOKIE_LOGIN']) && !empty($_COOKIE['COOKIE_LOGIN']))
setcookie('COOKIE_LOGIN',$objUserSerialized,time() - 86400 );
PHP converts unknown literals to strings and throws a warning. Your php.ini probably had the error reporting level to not display warnings but someone may have updated it or something. In either case, it is bad practice to take advantange of PHP's behavior in this case.
For more information, check out the php documentation:
Always use quotes around a string literal array index. For example, $foo['bar'] is correct, while $foo[bar] is not.
This is wrong, but it works. The reason is that this code has an undefined constant (bar) rather than a string ('bar' - notice the quotes). PHP may in future define constants which, unfortunately for such code, have the same name. It works because PHP automatically converts a bare string (an unquoted string which does not correspond to any known symbol) into a string which contains the bare string. For instance, if there is no defined constant named bar, then PHP will substitute in the string 'bar' and use that.
You can't say $_COOKIE[COOKIE_LOGIN] without error unless COOKIE_LOGIN is an actual constant that you have defined, e.g.:
define("COOKIE_LOGIN", 5);
Some people have habits where they will write code like:
$ary[example] = 5;
echo $ary[example];
Assuming that "example" (as a string) will be the array key. PHP has in the past excused this behavior, if you disable error reporting. It's wrong, though. You should be using $_COOKIE["COOKIE_LOGIN"] unless you have explicitly defined COOKIE_LOGIN as a constant.