Rishiraj Bayerd

Hi All,
With due respect to all , can someone advice me how to start learning Artificial Intelligence (AI). I am very new in AI and want to start from the basics. I am very grateful, if you also mention the prerequisite (if any), any resources like books, websites etc.

Thanks & Regards
Rishi

It is do-nothing situation with if block.
The first snippet just act as an instruction. That is, the execution of just immediate instruction of if
never depends it.

The second snippet is an if -block with empty scope. So, there are nothing to execute after result of the if-condition become true.

So, the result of the snippets become same.

Hi,
I am trying to develop a simple ejb 3.0 application (just select a row from the database) using Netbeans 6.7. But I cannot understand how to configure the variours resources (like JNDI,Connection Pool etc.) of GlassFish 2.1 Application Server I am using Oracle10g as database.

Anybody please help me...

regards
Rishi

Hi,
I have two question....

1.I want to know how garbage collector works on mutable objects(such
asobject of String & Wrapper class) ?

2.For knowing the length of array, we use array.length.... my quetion is what is length ?

thanks

with regards
Rishi

Hi Shelly,
error 1: you cannot write a method with in another methor(here main()).
error 2: Let's consider you declear call() in the class shelly,when you invoke call() method,you should placed a semicolon( at the end of the line of invocation(i.e call() .

But here main probleam is methed declare within another method.

regards
Rishi

Hi Neha,
i think it does not depends on array type(i.e. Anonymous,Simple). All mistry hide at first foreach loop. I try to draw the picture.

Original code:
for(int i: arr1) arr1[i] = 0;

int i:arr1 means the value of the i is the first element of arr1(0th place of arr1) i.e. the value of the i is 1.

For fisrt time... Then arr1[i]=0 becomes arr1[1]=0 .
So,(Original array) {1,2,3,4}=>{1,0,3,4}

Again for second time...
The value of the i is the second element of arr1(1st place of arr1) i.e. the value of the i is 0. Then arr1[i]=0 becomes arr1[0]=0
So,(Original array) {1,0,3,4}=>{0,0,3,4}


Again for Third time...
The value of the i is the third element of arr1(2nd place of arr1) i.e. the value of the i is 3. Then arr1[i]=0 becomes arr1[3]=0
So,(Original array) {0,0,3,4}=>{0,0,3,0}

Again for last time...(length of arr1 is 4)
The value of the i is the fourth element of arr1(3rd place of arr1) i.e. the value of the i is 0. Then arr1[i]=0 becomes arr1[0]=0 So,(Original array) {0,0,3,0}=>{0,0,3,0}

So the final output will be 0030

thanks....

with regards

Rishi

Hi Al Mamun,
Excellent explanation. Simply superb� if we take int [] arr={10,20,30,40} instead of int [] arr={1,2,3,4} we got ArrayOutofBound exception , because at for(i:arr), the value of is 10 i.e value of first element(value 0th place) and that becomes use as index of arr at line arr[i]=0;

with regards
Rishi

I think this program does't complile becouse of parameter of 2nd & 3rd checkThis method.The comlpiler does't detirmine which one will be called.

Main point to remember.... (from K & B book)
Widening beats boxing
Widening beats var-args

with regards
Rishi

I think this program does't complile becouse of parameter of 2nd & 3rd checkThis method.The comlpiler does't detirmine which one will be called.

Main point to remember.... (from K & B book)
Widening beats boxing
Widening beats var-args