get random number

meng

Hi,
I am trying to generate two different random numbers to display two different dice numbers on two image controls. The btn click event had the following code(below), every time I click the btn click event, both image controls show the same dice number. Please advise how to display different dice numbers on each image control?
Example, imgDDiceOne.Image will show dice number 5, imgDDiceOne.Image will show dice number 1.
Random rd1 = new Random();
Random rd2 = new Random();

int rol1 = rd1.Next(1, 6);
int rol2 = rd2.Next(1, 6);

imgDDiceOne.Image = Image.FromFile(filepath + "die-" + rol1.ToString() + ".gif");

imgDDiceTwo.Image = Image.FromFile(filepath + "die-" + rol2.ToString() + ".gif");

thanks

5 Replies

Know the answer? Post it — somebody with the same question will find it here.

Vulpes14y ago
Accepted answer
Incidentally, you should be using:

int rol1 = rd1.Next(1, 7);
int rol2 = rd2.Next(1, 7);

to get a random number from 1 to 6, as the upper bound of the Next method is exclusive.
meng14y ago
thank you, it works!
Vulpes14y ago
I suspect the problem is that the the two Random objects, which are being created in quick succession, are being initialized with the same (time dependent) seed and are therefore producing the same random numbers.

Try changing this line:

Random rd2 = new Random();

to this:

Random rd2 = new Random(Int32.MaxValue/2);

I'd also consider making the two Random variables, rd1 and rd2, into private fields (i.e. form level variables) rather than local variables. If you do that, then you won't be creating new objects each time the dice are rolled.
meng14y ago
i changed the second one to rol2.ToString() after posted the code, after the change, it's still showing the same result - double same images. thanks
Scott Lysle14y ago
It looks like you are setting both images using rol1.ToString(). Looks like if you change the second to rol2.ToString() you'll be in business.