PHP Incorrect variable declaration
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
asked Nov 16 '18 at 16:02
itsliamocoitsliamoco
295217
|
show 2 more comments
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 $variable is initially declared as a string.
But why does this work in PHP 5.6? What can I do to make this work in 7.2 without trawling through thousands of lines of code?
Is there a directive like strict_types I can use?
php php-5.6 legacy-code php-7.2
asked Nov 16 '18 at 16:02
itsliamocoitsliamoco
295217
1
For anyone curious about the output in different PHP versions: 3v4l.org/Q8iiY
– Namoshek
Nov 16 '18 at 16:07
4
It's because the string'key'when treated as an integer becomes0- since$variableis a string what you're doing is'Hello World'[0]... which isH
– 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,phpstandoesn'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
|
show 2 more comments
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
asked Nov 16 '18 at 16:02
itsliamocoitsliamoco
295217
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 $variable is initially declared as a string.
But why does this work in PHP 5.6? What can I do to make this work in 7.2 without trawling through thousands of lines of code?
Is there a directive like strict_types I can use?
php php-5.6 legacy-code php-7.2
php php-5.6 legacy-code php-7.2
asked Nov 16 '18 at 16:02
itsliamocoitsliamoco
295217
asked Nov 16 '18 at 16:02
itsliamocoitsliamoco
295217
edited Nov 16 '18 at 18:08
itsliamoco
asked Nov 16 '18 at 16:02
itsliamocoitsliamoco
295217
asked Nov 16 '18 at 16:02
itsliamocoitsliamoco
295217
asked Nov 16 '18 at 16:02
itsliamocoitsliamoco
295217
295217
1
For anyone curious about the output in different PHP versions: 3v4l.org/Q8iiY
– Namoshek
Nov 16 '18 at 16:07
4
It's because the string'key'when treated as an integer becomes0- since$variableis a string what you're doing is'Hello World'[0]... which isH
– 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,phpstandoesn'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
|
show 2 more comments
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 becomes0- since$variableis a string what you're doing is'Hello World'[0]... which isH
– CD001
Nov 16 '18 at 16:07
1
^^^ So remove$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,phpstandoesn'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
|
show 2 more comments
1 Answer
1
active
oldest
votes
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.
answered Nov 16 '18 at 16:12
Sander VisserSander Visser
2,75011934
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53341454%2fphp-incorrect-variable-declaration%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered Nov 16 '18 at 16:12
Sander VisserSander Visser
2,75011934
add a comment |
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.
answered Nov 16 '18 at 16:12
Sander VisserSander Visser
2,75011934
add a comment |
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.
answered Nov 16 '18 at 16:12
Sander VisserSander Visser
2,75011934
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.
answered Nov 16 '18 at 16:12
Sander VisserSander Visser
2,75011934
edited Nov 16 '18 at 17:05
answered Nov 16 '18 at 16:12
Sander VisserSander Visser
2,75011934
answered Nov 16 '18 at 16:12
Sander VisserSander Visser
2,75011934
answered Nov 16 '18 at 16:12
Sander VisserSander Visser
2,75011934
2,75011934
add a comment |
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53341454%2fphp-incorrect-variable-declaration%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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 becomes0- since$variableis a string what you're doing is'Hello World'[0]... which isH– 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,
phpstandoesn'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