Creating Multiple Objects?

[Dante Hawke]

Hey all,

I need to make 100 'Customer' Objects that are each assigned a randomly decided on string as their parameter. - I can do this via a random number generator with each number referring to a different string.

e.g.

Possible strings: "Hello" or "Goodbye" or "Farewell".

Customer C1 = new Customer("Hello"); x 100

How can I guarantee that the reference variable to each new object will be different? As after I create 100 of them, I must call a method on each of them 100 times.

Any help is most appreciated,

Many Thanks.

Customer P1 = new Customer("Hello");
Customer P2 = new Customer("Goodbye");
Customer P3 = new Customer("Farewell");
				
C1.queueUp(Q1);
C2.queueUp(Q1);
C3.queueUp(Q1);
				
		/*for(int i=0; i<99; i++) {
			ranNum = (generator.nextInt(3) + 1);
			
			if (ranNum == 1) {
				Customer Ci = new Customer("Hello");
				Ci.queueUp(Q1);
			}
			
			if (ranNum == 2) {
				Customer Ci = new Customer("Goodbye");
				Ci.queueUp(Q1);
			}
			
			if (ranNum == 3) {
				Customer Ci = new Customer("Farewell");
				Ci.queueUp(Q1);
			}				
		}*/

[David Newton]

If you need to keep a reference to each object then you'll want to use an array or some sort of collection.

[Dante Hawke]

Ah of course how silly of me. I now create an array and create each object with a reference variable and place that within each element. Does it matter that the reference variable is the same still though?

Random generator = new Random();
int ranNum = 0;
Customer[] customers = new Customer[100];


for(int i=0; i<100; i++) {
			
			ranNum = (generator.nextInt(3) + 1);
			
			if (ranNum == 1) {
				Customer Ci = new Customer("Hello");
				customers[i] = Ci;
				customers[i].queueUp(Q1);
			}
			
			if (ranNum == 2) {
				Customer Ci = new Customer("Goodbye");
				customers[i] = Ci;
				customers[i].queueUp(Q1);
			}
			
			if (ranNum == 3) {
				Customer Ci = new Customer("Farewell");
				customers[i] = Ci;
				customers[i].queueUp(Q1);
			}				
		}

[Campbell Ritchie]

You want an array which contains "Hello" "Goodbye" and "Farewell" and use the random number as the index to use those numbers in the Customer constructor.

[fred rosenberger]

i would also factor out the common code, and combine the three 'if''s into a single if-else. Note that if i ==1, it cannot == 2 or ==3. you'd be better off with

if (ranNum == 1) {}
else if (ranNum == 2) {}
else if (ranNum == 3) {}

further, it's redundant to have

customers[i] = Ci;
customers[i].queueUp(Q1);

three times. I'd factor it out to outside of your if-else entirely.

[Rob Spoor]

fred rosenberger wrote:Note that if i ==1, it cannot == 2 or ==3. you'd be better off with

if (ranNum == 1) {}
else if (ranNum == 2) {}
else if (ranNum == 3) {}

Or a switch statement:

switch (ranNum)
{
    case 1:
        // code
        break;
    case 2:
        // code
        break;
    case 3:
        // code
        break;
}

[fred rosenberger]

Good point...I, personally, don't like switch statements, but have no good reason for my dislike. I can't say there is anything wrong with them...they're just not my style.

[Dante Hawke]

Thanks for the input guys, my final 'working' code is:

int ranNum = 0;
Customer[] customers = new Customer[100];
String[] phrase = {"Hello", "Goodbye", "Farewell"};

for(int i=0; i<passengers.length; i++) {			
    ranNum = (int) (Math.random() * 3);
    customers[i] = new Customer(phrase[ranNum]);
    customers[i].queueUp(Q1);
}

[Campbell Ritchie]

That looks a lot better. Well done.

[Campbell Ritchie]

But you would do well to get your spacing consistent. You have some binary operators with spaces and some without. That will lose you marks. Recommend: a binary operator is preceded and followed by single spaces. That includes =.