Re: [PHP] No notices for undefined index
On Fri, 2010-04-09 at 14:15 +0100, Nathan Rixham wrote: > Ashley Sheridan wrote: > > can't find anything in the manual that explains what should happen when > > you treat a string like an array in PHP. > > http://www.php.net/manual/en/language.types.string.php#language.types.string.substr > > :) > Thanks, and I think that page answers the whole question: "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. Negative offset emits E_NOTICE in write but reads empty string. Only the first character of an assigned string is used. Assigning empty string assigns NUL byte." Thanks, Ash http://www.ashleysheridan.co.uk
Re: [PHP] No notices for undefined index
Ashley Sheridan wrote: > can't find anything in the manual that explains what should happen when > you treat a string like an array in PHP. http://www.php.net/manual/en/language.types.string.php#language.types.string.substr :) -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] No notices for undefined index
From: Shawn McKenzie > Bob McConnell wrote: >> In the first case, $a=5 creates a multi-typed variable. The interpreter >> makes its best guess how the next two expressions should be interpreted. >> In both cases, they look a lot like an index into a character array >> (string), and 'test' evaluates numerically to zero. Both are valid >> offsets for a string, so no messages are generated. >> >> In the second case, $a is explicitly declared as an array. This give the >> interpreter a lot more detail to work from. The two expressions are now >> an index and a key for the array. But both of them evaluate to offsets >> that have not been assigned, which raises a flag and creates the >> warnings. >> >> Such are the joys of loosely typed languages. > > Yes, this is what I was thinking as well, however: > > $a=5; > print $a[0]; // if it is index 0 then it should print 5 yes? > print $a[100]; // there is no index 100 so why no notice? I'm assuming that the PHP interpreter works much like a C compiler. i.e. It doesn't keep track of the size of strings. It knows that $a maps to a memory location, and $a[100] maps to that location plus 100 characters. As long as that is still a valid memory address for this process, it doesn't see anything wrong. If it is outside the process memory, you are more likely to get a General Protection Fault, or the equivalent OS error. In security parlance, this is what is known as a buffer overflow error. The application programmer is responsible for keeping track of string sizes and insuring that indexes don't move past the end of the allocated space. It is also why functions like snprintf should be used instead of sprintf. Bob McConnell -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] No notices for undefined index
On Fri, 2010-04-09 at 07:52 +0530, kranthi wrote: > >> print $a[0]; // prints 5 > >> print $a[100]; // Notice: Uninitialized string offset: 100 > Yup, this should happen when 5 is treated as an array of characters. > In other words as a string. > $a = '5'; > echo $a[0]; > echo $a[100]; > gives you the expected result > > regarding the original question, i think that the interpreter is > prefilling the variable with null > > $a = 5; > var_dump(isset($a[0])); > var_dump($a[0]); > > since $a[0] is already assigned (to null) the interpreter is not > throwing a notice > > $b = null; > var_dump(isset($b)); > var_dump($b); > If $a is treated as an array of characters, then $a[0] is always the first character, not null. Thanks, Ash http://www.ashleysheridan.co.uk
Re: [PHP] No notices for undefined index
Hello Shawn, Why dont you report a bug? When we know the expected behavior or the way it SHOULD behave. and its not behaving that way. Its certainly a bug.. Only then we can know the real reason why the novicas are not showing up. On 4/8/10, Shawn McKenzie wrote: > So the first two print statements generate NO notices, while the second > obviously generates: > > Notice: Undefined offset: 1 in /home/shawn/www/test.php on line 11 > > Notice: Undefined index: test in /home/shawn/www/test.php on line 12 > > This sucks. A bug??? > > error_reporting(E_ALL); > ini_set('display_errors', '1'); > > > $a = 5; > print $a[1]; > print $a['test']; > > $a = array(); > print $a[1]; > print $a['test']; > > -- > Thanks! > -Shawn > http://www.spidean.com > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > > -- Sent from my mobile device Shiplu Mokaddim My talks, http://talk.cmyweb.net Follow me, http://twitter.com/shiplu SUST Programmers, http://groups.google.com/group/p2psust Innovation distinguishes bet ... ... (ask Steve Jobs the rest) -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] No notices for undefined index
>> print $a[0]; // prints 5 >> print $a[100]; // Notice: Uninitialized string offset: 100 Yup, this should happen when 5 is treated as an array of characters. In other words as a string. $a = '5'; echo $a[0]; echo $a[100]; gives you the expected result regarding the original question, i think that the interpreter is prefilling the variable with null $a = 5; var_dump(isset($a[0])); var_dump($a[0]); since $a[0] is already assigned (to null) the interpreter is not throwing a notice $b = null; var_dump(isset($b)); var_dump($b); -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] No notices for undefined index
On Thu, 2010-04-08 at 15:22 -0500, Shawn McKenzie wrote: > Shawn McKenzie wrote: > > Bob McConnell wrote: > >> In the first case, $a=5 creates a multi-typed variable. The interpreter > >> makes its best guess how the next two expressions should be interpreted. > >> In both cases, they look a lot like an index into a character array > >> (string), and 'test' evaluates numerically to zero. Both are valid > >> offsets for a string, so no messages are generated. > >> > >> In the second case, $a is explicitly declared as an array. This give the > >> interpreter a lot more detail to work from. The two expressions are now > >> an index and a key for the array. But both of them evaluate to offsets > >> that have not been assigned, which raises a flag and creates the > >> warnings. > >> > >> Such are the joys of loosely typed languages. > >> > >> Bob McConnell > > > > Yes, this is what I was thinking as well, however: > > > > $a=5; > > print $a[0]; // if it is index 0 then it should print 5 yes? > > print $a[100]; // there is no index 100 so why no notice? > > > > $a='5'; > print $a[0]; // prints 5 > print $a[100]; // Notice: Uninitialized string offset: 100 > > So it seems, in the first case with the integer 5 that the interpreter > is saying: > > - Since $a is not an array I'll treat $a[0] and $a[100] as a string > offset, but since $a is not a string I won't do anything. > > Just seems stupid IMHO. > > -- > Thanks! > -Shawn > http://www.spidean.com > I think it just returns null if the offset goes beyond the length of the string. In C and C++ doing something like this would take you beyond that variables memory allocation into neighbouring variables. I believe PHP is trying to prevent problems where that might occur by returning null instead. This is only conjecture as I don't know exactly what happens, and I can't find anything in the manual that explains what should happen when you treat a string like an array in PHP. However, throwing some sort of error or notice would be nice, but could be worked around by checking the array type and length (the count() function returns the string length I believe as well as the array size) as it does seem that PHP is treating a string like a special sort of array. Thanks, Ash http://www.ashleysheridan.co.uk
Re: [PHP] No notices for undefined index
Shawn McKenzie wrote: > Bob McConnell wrote: >> In the first case, $a=5 creates a multi-typed variable. The interpreter >> makes its best guess how the next two expressions should be interpreted. >> In both cases, they look a lot like an index into a character array >> (string), and 'test' evaluates numerically to zero. Both are valid >> offsets for a string, so no messages are generated. >> >> In the second case, $a is explicitly declared as an array. This give the >> interpreter a lot more detail to work from. The two expressions are now >> an index and a key for the array. But both of them evaluate to offsets >> that have not been assigned, which raises a flag and creates the >> warnings. >> >> Such are the joys of loosely typed languages. >> >> Bob McConnell > > Yes, this is what I was thinking as well, however: > > $a=5; > print $a[0]; // if it is index 0 then it should print 5 yes? > print $a[100]; // there is no index 100 so why no notice? > $a='5'; print $a[0]; // prints 5 print $a[100]; // Notice: Uninitialized string offset: 100 So it seems, in the first case with the integer 5 that the interpreter is saying: - Since $a is not an array I'll treat $a[0] and $a[100] as a string offset, but since $a is not a string I won't do anything. Just seems stupid IMHO. -- Thanks! -Shawn http://www.spidean.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] No notices for undefined index
Bob McConnell wrote: > In the first case, $a=5 creates a multi-typed variable. The interpreter > makes its best guess how the next two expressions should be interpreted. > In both cases, they look a lot like an index into a character array > (string), and 'test' evaluates numerically to zero. Both are valid > offsets for a string, so no messages are generated. > > In the second case, $a is explicitly declared as an array. This give the > interpreter a lot more detail to work from. The two expressions are now > an index and a key for the array. But both of them evaluate to offsets > that have not been assigned, which raises a flag and creates the > warnings. > > Such are the joys of loosely typed languages. > > Bob McConnell Yes, this is what I was thinking as well, however: $a=5; print $a[0]; // if it is index 0 then it should print 5 yes? print $a[100]; // there is no index 100 so why no notice? -- Thanks! -Shawn http://www.spidean.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] No notices for undefined index
Andre Polykanine wrote: > Hello Shawn, > > Hm... isn't it expected behavior? Since you haven't defined a > $a['test'] item, PHP throws a notice... or I'm wrong? Yes it is expected. I'm saying the opposite that it doesn't in the first case. -- Thanks! -Shawn http://www.spidean.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] No notices for undefined index
In the first case, $a=5 creates a multi-typed variable. The interpreter makes its best guess how the next two expressions should be interpreted. In both cases, they look a lot like an index into a character array (string), and 'test' evaluates numerically to zero. Both are valid offsets for a string, so no messages are generated. In the second case, $a is explicitly declared as an array. This give the interpreter a lot more detail to work from. The two expressions are now an index and a key for the array. But both of them evaluate to offsets that have not been assigned, which raises a flag and creates the warnings. Such are the joys of loosely typed languages. Bob McConnell -Original Message- From: Andre Polykanine [mailto:an...@oire.org] Sent: Thursday, April 08, 2010 1:45 PM To: Shawn McKenzie Cc: php-general@lists.php.net Subject: Re: [PHP] No notices for undefined index Hello Shawn, Hm... isn't it expected behavior? Since you haven't defined a $a['test'] item, PHP throws a notice... or I'm wrong? -- With best regards from Ukraine, Andre Skype: Francophile; Wlm&MSN: arthaelon @ yandex.ru; Jabber: arthaelon @ jabber.org Yahoo! messenger: andre.polykanine; ICQ: 191749952 Twitter: m_elensule - Original message - From: Shawn McKenzie To: php-general@lists.php.net Date: Thursday, April 8, 2010, 8:36:21 PM Subject: [PHP] No notices for undefined index So the first two print statements generate NO notices, while the second obviously generates: Notice: Undefined offset: 1 in /home/shawn/www/test.php on line 11 Notice: Undefined index: test in /home/shawn/www/test.php on line 12 This sucks. A bug??? error_reporting(E_ALL); ini_set('display_errors', '1'); $a = 5; print $a[1]; print $a['test']; $a = array(); print $a[1]; print $a['test']; -- Thanks! -Shawn http://www.spidean.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] No notices for undefined index
Hello Shawn, Hm... isn't it expected behavior? Since you haven't defined a $a['test'] item, PHP throws a notice... or I'm wrong? -- With best regards from Ukraine, Andre Skype: Francophile; Wlm&MSN: arthaelon @ yandex.ru; Jabber: arthaelon @ jabber.org Yahoo! messenger: andre.polykanine; ICQ: 191749952 Twitter: m_elensule - Original message - From: Shawn McKenzie To: php-general@lists.php.net Date: Thursday, April 8, 2010, 8:36:21 PM Subject: [PHP] No notices for undefined index So the first two print statements generate NO notices, while the second obviously generates: Notice: Undefined offset: 1 in /home/shawn/www/test.php on line 11 Notice: Undefined index: test in /home/shawn/www/test.php on line 12 This sucks. A bug??? error_reporting(E_ALL); ini_set('display_errors', '1'); $a = 5; print $a[1]; print $a['test']; $a = array(); print $a[1]; print $a['test']; -- Thanks! -Shawn http://www.spidean.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] No notices for undefined index
On Thu, 2010-04-08 at 12:36 -0500, Shawn McKenzie wrote: > So the first two print statements generate NO notices, while the second > obviously generates: > > Notice: Undefined offset: 1 in /home/shawn/www/test.php on line 11 > > Notice: Undefined index: test in /home/shawn/www/test.php on line 12 > > This sucks. A bug??? > > error_reporting(E_ALL); > ini_set('display_errors', '1'); > > > $a = 5; > print $a[1]; > print $a['test']; > > $a = array(); > print $a[1]; > print $a['test']; > > -- > Thanks! > -Shawn > http://www.spidean.com > I think this goes back to the C style strings, where a string is just a collection of characters. I've noticed that in PHP you can treat a string as if it were an array of characters, so I guess in both cases above, it would be trying to return the second character, which is the termination character or a chr(0). In the second example, you've explicitely declared $a to be an array, so PHP creates a proper index for it, and then when you ask for an element that is not in that index list, it throws a notice at you. Thanks, Ash http://www.ashleysheridan.co.uk
[PHP] No notices for undefined index
So the first two print statements generate NO notices, while the second obviously generates: Notice: Undefined offset: 1 in /home/shawn/www/test.php on line 11 Notice: Undefined index: test in /home/shawn/www/test.php on line 12 This sucks. A bug??? error_reporting(E_ALL); ini_set('display_errors', '1'); $a = 5; print $a[1]; print $a['test']; $a = array(); print $a[1]; print $a['test']; -- Thanks! -Shawn http://www.spidean.com -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php