Query regarding random function

I am doing a random function this is what i hv done
static Random svcRand;
a=svcRand.nextInt();
its working perfectly but every time a random number is generated its a huge number......How can I keep it within a limit say between 1 and 100 always

showstopper wrote:
I am doing a random function this is what i hv done
static Random svcRand;
a=svcRand.nextInt(); This code would generate a NullPointerException, but I guess you initialize this somewhere.
>
its working perfectly but every time a random number is generated its a huge number......How can I keep it within a limit say between 1 and 100 alwaysThere are two ways
1. Easy: a = svcRand.nextInt(100) + 1; // number between 1 and 100
2. More general:
double start = ...;
double span = ...;
double val = svcRand.nextDouble() * span + start;

Similar Messages

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    Hi all.
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    Ok, I will clear myself with the following example :
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    Now, first my-thoughts; kindly correct me if I am wrong :
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    Kindly correct me I am wrong anywhere in the above.
    b) In case 2, we have a non-static function, synchronized on a static object. Here, again if obj1, and obj2 happen to call the function at the same time, obj1 will try to obtain lock for obj1.a; while obj2 will try to obtain lock for obj2.a. However, since obj1.a and obj2.a are the same, thus we will indeed obtain sychronisation.
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    d) In case 4, we have a static function, synchronized on a static object.
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    d) In case 4, we have a static function, synchronized on a static object. Here, again if obj1, and obj2 happen to call the function at the same time, obj1 will try to obtain lock for obj1.a; while obj2 will try to obtain lock for obj2.a.Again, obj1/2 don't call methods or try to obtain locks. The two different threads do that. And again, you mean First.b. obj1.b and obj2.b are the same as First.b. Does that make it clearer?
    Now, I have a query : what happens if the call is made in a classically static manner, i.e. using the statement "First.func_four;".That's what happens in any case whether you write obj1.func_four(), obj2.func)four(), or First.func_four(). All these are identical when func_four(0 is static.
    Now, consider this, suppose we have the same reference obj1, in two threads, and the call "obj1.func_four;" happens to occur at the same time from each of these threads. Thus, we have obj1 rying to obtain lock for obj1.aNo we don't, we have a thread trying to obtain the lock on First.b.
    and again obj1 trying to obtain lock for obj1.aYou mean obj2 and First.b, but obj2 doesn't obtain the lock, the thread does.
    which are the same locks. So, if obj1.a of the first thread obtains the lock, then it will enter the function no-doubt, but the call from the second thread will also succeed.Of course it won't. Your reasoning here makes zero sense..Once First.b is locked it is locked. End of story.
    Thus, effectively, our synchronisation is broken.No it isn't. The second thread will wait on the same First.b object that the first thread has locked.
    However in any case you have a much bigger problem here. You're autoboxing your local 'int' variable to a possibly brand-new Integer object every call, so there may be no synchronization at all.
    You need:
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    Hello
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    CORE     10.2.0.4.0     Production
    TNS for Solaris: Version 10.2.0.4.0 - Production
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    FROM
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                           1                     
    0                      1                     
    1                      1                     
    2                      2                     
    3                      3                     

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    Hi All,
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    Message was edited by:
    Zorry

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