As semicolon is there after if statement it does not have any effect on the program. It will remain as it is in the program and also will not not give any compiling and runtime error. Because the if statement get terminated due to semicolon.
[ February 29, 2008: Message edited by: sandhi mridul ] [ February 29, 2008: Message edited by: sandhi mridul ]
The is just a blank if statement which does nothing. So when is called it calls the test method, does nothing [if(false);] and then sample() method outputs the "Sample". Here sample() method is not conditional,it executes always irrespective of the flag.
Joined: Feb 24, 2008
Thanks guys that helped.
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