if (!$foo && !bar) actually won't gave error when $bar is typed as bar

the statement

if (!$foo && !$bar) { echo 'hi'; }

is typed as

if (!$foo && !bar) { echo 'hi'; }

(with the '$bar' mistyped as 'bar')

and i just found that PHP won't give error... is it true that the word bar is taken as a string literal... it seems a bit weird and sometimes it can give hard to catch error. many thanks.

 

error_reporting(E_ERROR | E_WARNING | E_PARSE | E_NOTICE);

PHP:

 

ah i see... so with error_reporting(E_ERROR | E_WARNING | E_PARSE | E_NOTICE);

it will say bar is undefined constant and will treat it as a string literal
(like if i do a var_dump(bar); )

 

Someone suggested using

error_reporting(E_ALL);

I have some code to tell what E_ALL includes... if someone knows of better ways of doing it and like to share could you post here. thanks.

$arrayErrorLevelName = array(
	'E_ERROR',
	'E_WARNING',
	'E_PARSE', 
	'E_NOTICE', 
	'E_CORE_ERROR',
	'E_CORE_WARNING', 
	'E_COMPILE_ERROR', 
	'E_COMPILE_WARNING', 
	'E_USER_ERROR', 
	'E_USER_WARNING', 
	'E_USER_NOTICE',
	'E_STRICT',
	'E_RECOVERABLE_ERROR',
	'E_DEPRECATED',
	'E_USER_DEPRECATED'
);
	
echo "<pre>On PHP " . phpversion() . "\nE_ALL includes\n--------------\n";
foreach ($arrayErrorLevelName as $errorLevelName) {
    eval("if ($errorLevelName & E_ALL) echo(\"$errorLevelName\n\");");    
}

PHP: