About Question enthuware.ocajp.i.v7.2.1242 :

Michailangelo · Post by **Michailangelo** » Mon Sep 24, 2012 9:01 am

I don't see the args[2] reference in the code.

Post by **admin** » Mon Sep 24, 2012 9:09 am

int i = Integer.parseInt(args[1]);   
System.out.println(args[i]);

args[1] is 2
so i is 2
therefore, args is args[2]

HTH,
Paul.

Michailangelo · Post by **Michailangelo** » Mon Sep 24, 2012 9:29 am

I should pay more attention

.

satar · Post by **satar** » Wed Feb 13, 2013 10:16 am

Yeah, this one got me also

deepa.patre · Post by **deepa.patre** » Tue Jun 18, 2013 11:00 am

I have a basic question,
why is the method name just parseInt() instead of parseInteger()?

Please reply,

Post by **admin** » Tue Jun 18, 2013 11:13 am

You have to ask Java designers that

peckw1 · Post by **peckw1** » Wed Apr 15, 2015 7:18 am

If I run the code with 4 and 5 I also get the exception, in this case shouldn't args[0] = 4 and args[1]=5?

Post by **admin** » Wed Apr 15, 2015 10:32 am

peckw1 wrote:If I run the code with 4 and 5 I also get the exception, in this case shouldn't args[0] = 4 and args[1]=5?

Yes, args[0] = 4 and args[1]=5. But I am not really sure what is your point.
Paul.

maltine · Post by **maltine** » Sat Jul 11, 2020 4:31 pm

System.out.println(i);//2
System.out.println(args); //i=2, so, args[2] --> throw an exception
Yes, this one got me also....

Enthuware