PHP Incorrect variable declaration












9















Debugging legacy code and I have a strange issue. The legacy code is being moved to PHP 7.2. I don't know which version of PHP it was originally written for but it does work in PHP 5.6.



Below is my example of the problem...



$variable = '';
$variable['key'] = 'Hello World!';

echo $variable['key'] // H


When I echo $variable['key'] it only gets the first character from the value. I know now that it is because $variable is initially declared as a string.



But why does this work in PHP 5.6? What can I do to make this work in 7.2 without trawling through thousands of lines of code?



Is there a directive like strict_types I can use?










share|improve this question




















  • 1





    For anyone curious about the output in different PHP versions: 3v4l.org/Q8iiY

    – Namoshek
    Nov 16 '18 at 16:07






  • 4





    It's because the string 'key' when treated as an integer becomes 0 - since $variable is a string what you're doing is 'Hello World'[0] ... which is H

    – CD001
    Nov 16 '18 at 16:07






  • 1





    ^^^ So remove $variable = ''; or do $variable = ;

    – AbraCadaver
    Nov 16 '18 at 16:08













  • Can't find any reference to this behavior in my limited searches but chances are you'll just want to fix all the buggy code -- which, hopefully your project isn't too large. This is incorrect syntax either way, not really sure why PHP 5 would behave like that except: PHP. Can't find any related php.ini settings either.

    – Sheng Slogar
    Nov 16 '18 at 16:11











  • Unfortunately, phpstan doesn't consider this bad code. So I also don't see a good way of automatically fixing these issues.

    – Namoshek
    Nov 16 '18 at 16:14
















9















Debugging legacy code and I have a strange issue. The legacy code is being moved to PHP 7.2. I don't know which version of PHP it was originally written for but it does work in PHP 5.6.



Below is my example of the problem...



$variable = '';
$variable['key'] = 'Hello World!';

echo $variable['key'] // H


When I echo $variable['key'] it only gets the first character from the value. I know now that it is because $variable is initially declared as a string.



But why does this work in PHP 5.6? What can I do to make this work in 7.2 without trawling through thousands of lines of code?



Is there a directive like strict_types I can use?










share|improve this question




















  • 1





    For anyone curious about the output in different PHP versions: 3v4l.org/Q8iiY

    – Namoshek
    Nov 16 '18 at 16:07






  • 4





    It's because the string 'key' when treated as an integer becomes 0 - since $variable is a string what you're doing is 'Hello World'[0] ... which is H

    – CD001
    Nov 16 '18 at 16:07






  • 1





    ^^^ So remove $variable = ''; or do $variable = ;

    – AbraCadaver
    Nov 16 '18 at 16:08













  • Can't find any reference to this behavior in my limited searches but chances are you'll just want to fix all the buggy code -- which, hopefully your project isn't too large. This is incorrect syntax either way, not really sure why PHP 5 would behave like that except: PHP. Can't find any related php.ini settings either.

    – Sheng Slogar
    Nov 16 '18 at 16:11











  • Unfortunately, phpstan doesn't consider this bad code. So I also don't see a good way of automatically fixing these issues.

    – Namoshek
    Nov 16 '18 at 16:14














9












9








9


0






Debugging legacy code and I have a strange issue. The legacy code is being moved to PHP 7.2. I don't know which version of PHP it was originally written for but it does work in PHP 5.6.



Below is my example of the problem...



$variable = '';
$variable['key'] = 'Hello World!';

echo $variable['key'] // H


When I echo $variable['key'] it only gets the first character from the value. I know now that it is because $variable is initially declared as a string.



But why does this work in PHP 5.6? What can I do to make this work in 7.2 without trawling through thousands of lines of code?



Is there a directive like strict_types I can use?










share|improve this question
















Debugging legacy code and I have a strange issue. The legacy code is being moved to PHP 7.2. I don't know which version of PHP it was originally written for but it does work in PHP 5.6.



Below is my example of the problem...



$variable = '';
$variable['key'] = 'Hello World!';

echo $variable['key'] // H


When I echo $variable['key'] it only gets the first character from the value. I know now that it is because $variable is initially declared as a string.



But why does this work in PHP 5.6? What can I do to make this work in 7.2 without trawling through thousands of lines of code?



Is there a directive like strict_types I can use?







php php-5.6 legacy-code php-7.2






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 16 '18 at 18:08







itsliamoco

















asked Nov 16 '18 at 16:02









itsliamocoitsliamoco

295217




295217








  • 1





    For anyone curious about the output in different PHP versions: 3v4l.org/Q8iiY

    – Namoshek
    Nov 16 '18 at 16:07






  • 4





    It's because the string 'key' when treated as an integer becomes 0 - since $variable is a string what you're doing is 'Hello World'[0] ... which is H

    – CD001
    Nov 16 '18 at 16:07






  • 1





    ^^^ So remove $variable = ''; or do $variable = ;

    – AbraCadaver
    Nov 16 '18 at 16:08













  • Can't find any reference to this behavior in my limited searches but chances are you'll just want to fix all the buggy code -- which, hopefully your project isn't too large. This is incorrect syntax either way, not really sure why PHP 5 would behave like that except: PHP. Can't find any related php.ini settings either.

    – Sheng Slogar
    Nov 16 '18 at 16:11











  • Unfortunately, phpstan doesn't consider this bad code. So I also don't see a good way of automatically fixing these issues.

    – Namoshek
    Nov 16 '18 at 16:14














  • 1





    For anyone curious about the output in different PHP versions: 3v4l.org/Q8iiY

    – Namoshek
    Nov 16 '18 at 16:07






  • 4





    It's because the string 'key' when treated as an integer becomes 0 - since $variable is a string what you're doing is 'Hello World'[0] ... which is H

    – CD001
    Nov 16 '18 at 16:07






  • 1





    ^^^ So remove $variable = ''; or do $variable = ;

    – AbraCadaver
    Nov 16 '18 at 16:08













  • Can't find any reference to this behavior in my limited searches but chances are you'll just want to fix all the buggy code -- which, hopefully your project isn't too large. This is incorrect syntax either way, not really sure why PHP 5 would behave like that except: PHP. Can't find any related php.ini settings either.

    – Sheng Slogar
    Nov 16 '18 at 16:11











  • Unfortunately, phpstan doesn't consider this bad code. So I also don't see a good way of automatically fixing these issues.

    – Namoshek
    Nov 16 '18 at 16:14








1




1





For anyone curious about the output in different PHP versions: 3v4l.org/Q8iiY

– Namoshek
Nov 16 '18 at 16:07





For anyone curious about the output in different PHP versions: 3v4l.org/Q8iiY

– Namoshek
Nov 16 '18 at 16:07




4




4





It's because the string 'key' when treated as an integer becomes 0 - since $variable is a string what you're doing is 'Hello World'[0] ... which is H

– CD001
Nov 16 '18 at 16:07





It's because the string 'key' when treated as an integer becomes 0 - since $variable is a string what you're doing is 'Hello World'[0] ... which is H

– CD001
Nov 16 '18 at 16:07




1




1





^^^ So remove $variable = ''; or do $variable = ;

– AbraCadaver
Nov 16 '18 at 16:08







^^^ So remove $variable = ''; or do $variable = ;

– AbraCadaver
Nov 16 '18 at 16:08















Can't find any reference to this behavior in my limited searches but chances are you'll just want to fix all the buggy code -- which, hopefully your project isn't too large. This is incorrect syntax either way, not really sure why PHP 5 would behave like that except: PHP. Can't find any related php.ini settings either.

– Sheng Slogar
Nov 16 '18 at 16:11





Can't find any reference to this behavior in my limited searches but chances are you'll just want to fix all the buggy code -- which, hopefully your project isn't too large. This is incorrect syntax either way, not really sure why PHP 5 would behave like that except: PHP. Can't find any related php.ini settings either.

– Sheng Slogar
Nov 16 '18 at 16:11













Unfortunately, phpstan doesn't consider this bad code. So I also don't see a good way of automatically fixing these issues.

– Namoshek
Nov 16 '18 at 16:14





Unfortunately, phpstan doesn't consider this bad code. So I also don't see a good way of automatically fixing these issues.

– Namoshek
Nov 16 '18 at 16:14












1 Answer
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oldest

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6














From php.net




Warning
Writing to an out of range offset pads the string with spaces. Non-integer types are converted to integer. Illegal offset type emits E_NOTICE. Only the first character of an assigned string is used. As of PHP 7.1.0, assigning an empty string throws a fatal error. Formerly, it assigned a NULL byte.




http://php.net/manual/en/language.types.string.php#language.types.string.substr



So "key" is converted to 0, and the first character is set.
Because this is a char type, only "H" is set from the given string.



$variable = '';
$variable['key'] = 'Hello World!';

echo $variable;
echo $variable['key'];


If you change your code to the above you can see better what happens.



So the text 'ello World!' is lost and gone in PHP >= 7.1 because you set the first character, the type stays string.



In php 5.6 you will get
Notice: Array to string conversion in /in/N2poP on line 6



So in prior versions you overwrite the complete variable, and the initial empty string would be gone, PHP simply creates a new array. This behavior only happens with an empty string!



This is also noted in the documentation:
http://php.net/manual/en/language.types.string.php#language.types.string.substr




Note: As of PHP 7.1.0, applying the empty index operator on an empty
string throws a fatal error. Formerly, the empty string was silently
converted to an array.




The easiest solution would be removing the $variable = ''; part, it's invalid anyway and never used in your legacy code. or by replacing it with $variable = ;



Because this behavior only happens with an empty string in php < 7.1 you could use a regular expression to find all places where you should refactor to fix the issue.






share|improve this answer

























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    From php.net




    Warning
    Writing to an out of range offset pads the string with spaces. Non-integer types are converted to integer. Illegal offset type emits E_NOTICE. Only the first character of an assigned string is used. As of PHP 7.1.0, assigning an empty string throws a fatal error. Formerly, it assigned a NULL byte.




    http://php.net/manual/en/language.types.string.php#language.types.string.substr



    So "key" is converted to 0, and the first character is set.
    Because this is a char type, only "H" is set from the given string.



    $variable = '';
    $variable['key'] = 'Hello World!';

    echo $variable;
    echo $variable['key'];


    If you change your code to the above you can see better what happens.



    So the text 'ello World!' is lost and gone in PHP >= 7.1 because you set the first character, the type stays string.



    In php 5.6 you will get
    Notice: Array to string conversion in /in/N2poP on line 6



    So in prior versions you overwrite the complete variable, and the initial empty string would be gone, PHP simply creates a new array. This behavior only happens with an empty string!



    This is also noted in the documentation:
    http://php.net/manual/en/language.types.string.php#language.types.string.substr




    Note: As of PHP 7.1.0, applying the empty index operator on an empty
    string throws a fatal error. Formerly, the empty string was silently
    converted to an array.




    The easiest solution would be removing the $variable = ''; part, it's invalid anyway and never used in your legacy code. or by replacing it with $variable = ;



    Because this behavior only happens with an empty string in php < 7.1 you could use a regular expression to find all places where you should refactor to fix the issue.






    share|improve this answer






























      6














      From php.net




      Warning
      Writing to an out of range offset pads the string with spaces. Non-integer types are converted to integer. Illegal offset type emits E_NOTICE. Only the first character of an assigned string is used. As of PHP 7.1.0, assigning an empty string throws a fatal error. Formerly, it assigned a NULL byte.




      http://php.net/manual/en/language.types.string.php#language.types.string.substr



      So "key" is converted to 0, and the first character is set.
      Because this is a char type, only "H" is set from the given string.



      $variable = '';
      $variable['key'] = 'Hello World!';

      echo $variable;
      echo $variable['key'];


      If you change your code to the above you can see better what happens.



      So the text 'ello World!' is lost and gone in PHP >= 7.1 because you set the first character, the type stays string.



      In php 5.6 you will get
      Notice: Array to string conversion in /in/N2poP on line 6



      So in prior versions you overwrite the complete variable, and the initial empty string would be gone, PHP simply creates a new array. This behavior only happens with an empty string!



      This is also noted in the documentation:
      http://php.net/manual/en/language.types.string.php#language.types.string.substr




      Note: As of PHP 7.1.0, applying the empty index operator on an empty
      string throws a fatal error. Formerly, the empty string was silently
      converted to an array.




      The easiest solution would be removing the $variable = ''; part, it's invalid anyway and never used in your legacy code. or by replacing it with $variable = ;



      Because this behavior only happens with an empty string in php < 7.1 you could use a regular expression to find all places where you should refactor to fix the issue.






      share|improve this answer




























        6












        6








        6







        From php.net




        Warning
        Writing to an out of range offset pads the string with spaces. Non-integer types are converted to integer. Illegal offset type emits E_NOTICE. Only the first character of an assigned string is used. As of PHP 7.1.0, assigning an empty string throws a fatal error. Formerly, it assigned a NULL byte.




        http://php.net/manual/en/language.types.string.php#language.types.string.substr



        So "key" is converted to 0, and the first character is set.
        Because this is a char type, only "H" is set from the given string.



        $variable = '';
        $variable['key'] = 'Hello World!';

        echo $variable;
        echo $variable['key'];


        If you change your code to the above you can see better what happens.



        So the text 'ello World!' is lost and gone in PHP >= 7.1 because you set the first character, the type stays string.



        In php 5.6 you will get
        Notice: Array to string conversion in /in/N2poP on line 6



        So in prior versions you overwrite the complete variable, and the initial empty string would be gone, PHP simply creates a new array. This behavior only happens with an empty string!



        This is also noted in the documentation:
        http://php.net/manual/en/language.types.string.php#language.types.string.substr




        Note: As of PHP 7.1.0, applying the empty index operator on an empty
        string throws a fatal error. Formerly, the empty string was silently
        converted to an array.




        The easiest solution would be removing the $variable = ''; part, it's invalid anyway and never used in your legacy code. or by replacing it with $variable = ;



        Because this behavior only happens with an empty string in php < 7.1 you could use a regular expression to find all places where you should refactor to fix the issue.






        share|improve this answer















        From php.net




        Warning
        Writing to an out of range offset pads the string with spaces. Non-integer types are converted to integer. Illegal offset type emits E_NOTICE. Only the first character of an assigned string is used. As of PHP 7.1.0, assigning an empty string throws a fatal error. Formerly, it assigned a NULL byte.




        http://php.net/manual/en/language.types.string.php#language.types.string.substr



        So "key" is converted to 0, and the first character is set.
        Because this is a char type, only "H" is set from the given string.



        $variable = '';
        $variable['key'] = 'Hello World!';

        echo $variable;
        echo $variable['key'];


        If you change your code to the above you can see better what happens.



        So the text 'ello World!' is lost and gone in PHP >= 7.1 because you set the first character, the type stays string.



        In php 5.6 you will get
        Notice: Array to string conversion in /in/N2poP on line 6



        So in prior versions you overwrite the complete variable, and the initial empty string would be gone, PHP simply creates a new array. This behavior only happens with an empty string!



        This is also noted in the documentation:
        http://php.net/manual/en/language.types.string.php#language.types.string.substr




        Note: As of PHP 7.1.0, applying the empty index operator on an empty
        string throws a fatal error. Formerly, the empty string was silently
        converted to an array.




        The easiest solution would be removing the $variable = ''; part, it's invalid anyway and never used in your legacy code. or by replacing it with $variable = ;



        Because this behavior only happens with an empty string in php < 7.1 you could use a regular expression to find all places where you should refactor to fix the issue.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Nov 16 '18 at 17:05

























        answered Nov 16 '18 at 16:12









        Sander VisserSander Visser

        2,75011934




        2,75011934






























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