java initializing array with random numbers

yyou commented out assignment

due to rounding


System.out.println("Random num: " + (int)(Math.random() * 9.0));

Try this:
int numItems = 20;
int[] array = new int[numItems];

for(int j = 0;j< array.length; j++){
 array[j] = (int)Math.random() * 9;
System.out.println("Random num: " + (int)Math.random() * 9);
}

@for_yan: I did that for testing - if you run it, the output is all zeros(???)

> (int)Math.random()

that will always round to zero

Yes rounding also true
but uncomment ykou need

But please post it correctly - it distracts

int numItems = 20;
int[] array = new int[numItems];

for(int j = 0;j< array.length; j++){
 array[j] = (int)(Math.random() * 9);
System.out.println("Random num: " + (int)Math.random() * 9);
}

int numItems = 20;
int[] array = new int[numItems];

for(int j = 0;j< array.length; j++){
 array[j] = (int)(Math.random() * 9);
System.out.println("Random num: " + array[j]);
}

@objects:  I thought it was a rounding problem, but it switched out 9 for 100, and it still gave me zeros.
What do I need to do to come up with random integers?

> @objects:  I thought it was a rounding problem, but it switched out 9 for 100, and it still gave me zeros.

the rounding has already occurred before that. its rounding the random to zero
try my first comment

http://helpdesk.objects.com.au/java/how-do-i-generate-a-random-integer-between-1-and-n-inclusive

Math.random()  generated=s from 0 to 1
when you cast it integer it always bcekl=omes zzero

Multiply firts then cast and it will be OK

If you cast a number between 0 and less than 1 to integer ut will use the floor
of that number - it will be zero

if you first multiply by 10 or by 100 then the floor will be integer and you can
cast and get integer number

That just repeats what I already posted??

> Thanks for your help - I used another set of () and that fixed it.

which is what I suggested in my first comment.
Why accept a comment that repeats what had already been posted?

> @objects:  I thought it was a rounding problem, but it switched out 9 for 100, and it still gave me zeros.

It's nothing to do with rounding:

System.out.println("Random num: " + (int)Math.random() * 9);

You cast the result of random() to an int, which will result in zero - game over.

What you need is the following, which will cast the result of random() * 9 to an int

System.out.println("Random num: " + (int)(Math.random() * 9));

Oh and it's nothing to do with widening/narrowing conversions either, which is why specifying 9.0 makes no difference to the below, since 9 will already be widened to a double automatically

( http:#35488970 )
>>
due to rounding


System.out.println("Random num: " + (int)(Math.random() * 9.0));
>>

The problem is due to

a. casting
b. precedence

The answer had been assigned correctly.

This is the code from  taken from  ID:35488981:

           int numItems = 20;
int[] array = new int[numItems];

for(int j = 0;j< array.length; j++){
 array[j] = (int)(Math.random() * 9);
System.out.println("Random num: " + array[j]);
} 

Select allOpen in new window

THis is the output of random numbers (by no means all zeros):

Random num: 5
Random num: 2
Random num: 4
Random num: 8
Random num: 0
Random num: 3
Random num: 3
Random num: 8
Random num: 4
Random num: 3
Random num: 5
Random num: 7
Random num: 8
Random num: 1
Random num: 1
Random num: 5
Random num: 6
Random num: 5
Random num: 0
Random num: 5

Select allOpen in new window

ID:35488999 also provided correct explanation

>>ID:35488999 also provided correct explanation

True - i actually missed that in all the noise ;)

Sorry for the confusion.  I missed the meaning of your earlier comment until you pointed it out.

Thanks to all for your help,
Lynn

Thanks Lynn :)