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Bug #43020 Warning message is missing with shuffle() more than one argument
Submitted: 2007-10-18 13:41 UTC Modified: 2007-10-18 14:39 UTC
From: nikhil dot gupta at in dot ibm dot com Assigned:
Status: Closed Package: Compile Warning
PHP Version: 5.2CVS-2007-10-18 (snap) OS: linux, windows
Private report: No CVE-ID: None
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From: nikhil dot gupta at in dot ibm dot com
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 [2007-10-18 13:41 UTC] nikhil dot gupta at in dot ibm dot com
When shuffle() function is called with two arguments (ie. one argument more than expected), it performs the operation of shuffling and donot throws warning message stating for more than expected number of arguments , as we see with other functions like end() etc.

Also when shuffle() is called with zero argument(ie. one less than minimum number of argument required) the warning message is not proper and do not say anything about the insufficient number of arguments.

The following can be added in the beginning of the function definition for PHP5 to check for the above scenarios:

if (ZEND_NUM_ARGS() != 1) {

Reproduce code:
$arr = array(1,2,3,4,5);
var_dump( shuffle() );
var_dump( shuffle($arr, 2) );

Expected result:
Warning: Wrong parameter count for shuffle() in %s on line %d

Warning: Wrong parameter count for shuffle() in %s on line %d

Actual result:
Warning: shuffle(): could not obtain parameters for parsing in %s on line %d


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 [2007-10-18 14:39 UTC]
This bug has been fixed in CVS.

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