[itachi]

If I start my server,server.log send this :

Код:

[08:31:48] sscanf warning: 'z' is deprecated, consider using 'S' instead.
[08:31:48] sscanf warning: No default value found.
[08:31:48] sscanf warning: Format specifier does not match parameter count.
[08:31:50] sscanf warning: 'z' is deprecated, consider using 'S' instead.
[08:31:50] sscanf warning: No default value found.
[08:31:50] sscanf warning: Strings without a length are deprecated, please add a destination size.

[X337]

You are using 'z' specifier for sscanf in your codes which is deprecated, replace them with 'S' if you have a default value, otherwise use 's' and also provide string destination length.

[AjaxM]

sscanf warning: 'z' is deprecated, consider using 'S' instead.

You are using 'z' somewhere in the script instead of using the specifier 's'.

Example:

PHP код:

if(sscanf(params, "z", ...)) 


Use 'S' instead of 'z'.

sscanf warning: Strings without a length are deprecated, please add a destination size.

Example:

PHP код:

new eg[5];
if(sscanf(params, "s", eg)) 


Should be

PHP код:

new eg[5];
if(sscanf(params, "s[5]", eg)) 


This can even cause your server to crash.

[iLearner]

Quote:

Originally Posted by AjaxM sscanf warning: 'z' is deprecated, consider using 'S' instead. You are using 'z' somewhere in the script instead of using the specifier 's'. Example: PHP код: if(sscanf(params, "z", ...))  Use 'S' instead of 'z'. sscanf warning: Strings without a length are deprecated, please add a destination size. Example: PHP код: new eg[5]; if(sscanf(params, "s", eg))  Should be PHP код: new eg[5]; if(sscanf(params, "s[5]", eg))  This can even cause your server to crash.

let's not exaggerate.