generate random number in java

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how can i generate random number in some range(for example 1-100) in java?

Posted 24-Jan-13 21:18pm

Coder93

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OriginalGriff · Accepted Answer · 2013-01-24T21:23:00

Solution 1

Try here: Java Practices -> Generate random numbers[^]

Posted 24-Jan-13 21:23pm

OriginalGriff

Comments

Coder93 25-Jan-13 5:36am

i use your solution , but it always generate same random number, what should i do ?

OriginalGriff 25-Jan-13 5:49am

Move the definition of randomGenerator outside the method - otherwise it is preset each time you call it.
If you look at the definition of Random it is preset to the system clock, so if you call the method 10 times in the same millisecond and create a new Random each time, you will get the same numbers.

Coder93 25-Jan-13 5:53am

I run the code after 20 min but, it generate the random number like before!!!

OriginalGriff 25-Jan-13 5:55am

Copy and paste the exact code you used.

Coder93 27-Jan-13 14:05pm

It is part of my code :

public class myclass {
private ……;
private Random random;
…….
public copkmeans (double [][] dataSet , int [][] constraint , int k, int i, int j , int l) throws IOException {
…….
random = new Random(i);
}
private void makeCluster (){
int randomNum;
int range = k;
while(range > 0){
randomNum = random.nextInt(i)+1;
System.out.println("random"+randomNum);
for(int m = 1; m < l ;m++){
if(constraint[m][1] == randomNum )
if(constraint[m][3] == 1)
if(clusters.contains(constraint[m][2])){
break;
}
if(constraint[m][2] == randomNum )
if(constraint[m][3] == 1)
if(clusters.contains(constraint[m][1])){
break;
}
clusters.add(randomNum);
used.add(randomNum);
myItem temp = new myItem(m, dataset , j);
temp.setHole();
clusterCenter.add(temp);
range--;
break;
}
}
int v = 1;
for (int a = 0 ; a < clusters.size() ; a++) {
partition[clusters.get(a)] = v;
v++;

}
}

OriginalGriff 28-Jan-13 3:55am

Well yes, it will. You are telling it to!
random = new Random(i);
Sets up a new random number generate using the same seed value each time. As a result, it generates the same sequence on numbers!
Take out the "i" and you should be fine:
random = new Random();

Coder93 30-Jan-13 4:47am

thank you ,
it work correctly.

OriginalGriff 30-Jan-13 5:34am

You're welcome!