ID: 17180
Updated by: [EMAIL PROTECTED]
Reported By: [EMAIL PROTECTED]
Status: Analyzed
Bug Type: Scripting Engine problem
PHP Version: 4.2.0
New Comment:
This behaviour is capable to confuse the developer and if this is
"features" it must be documented in manual.
Previous Comments:
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[2002-05-13 18:20:14] [EMAIL PROTECTED]
Well, but it's stupid to do something like that. It makes no sense to
assign anything to NOT(a variable), so PHP takes care of that by
changing the precedence a little in this case.
------------------------------------------------------------------------
[2002-05-13 17:56:54] [EMAIL PROTECTED]
Yes, I want ASSIGN value to $a and check assigned value.
But parser must say: "parser error", becouse it can not assign value to
constant.
Please reopen.
------------------------------------------------------------------------
[2002-05-13 17:48:54] [EMAIL PROTECTED]
"if (!$a = foo(FALSE))" --> you're assigning the output of foo(FALSE)
to $a
"if (!$a == foo(FALSE))" --> you're comparing !$a and foo(FALSE)
------------------------------------------------------------------------
[2002-05-13 09:36:32] [EMAIL PROTECTED]
Why & How this code will work?
<?
function foo($flag)
{
return $flag;
}
$a=TRUE;
echo "if (!\$a = foo(FALSE))) is ";
if (!$a = foo(FALSE))
echo "true";
else
echo "false";
echo "\n";
var_dump($a);
echo "\n";
?>
Output:
if (!$a = foo(FALSE))) is true
bool(false)
http://www.php.net/manual/en/language.operators.php "Operator
Precedence"
`!` has more precedence than `=`
And after `!` we must have boolean constant in left side:
FALSE = foo()
Explain to me pls that I do not understand
P.S. in C & Perl (!$a = foo()) is not valid expression
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Edit this bug report at http://bugs.php.net/?id=17180&edit=1
