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SSCANF WARNING - Printable Version

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+--- Thread: SSCANF WARNING (/showthread.php?tid=632832)



SSCANF WARNING - itachi - 20.04.2017

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.



Re: SSCANF WARNING - X337 - 20.04.2017

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.


Re: SSCANF WARNING - AjaxM - 20.04.2017

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.


Re: SSCANF WARNING - iLearner - 20.04.2017

Quote:
Originally Posted by AjaxM
Посмотреть сообщение
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.