I have such for loop and when step is (0;1) it becomes infinite. If step is [1;..) it works well.
public interface FindMinI {
double function(double x);
static double findMinOfFuncOnInterval(int begin, int end, double step, FindMinI func)
{
double min = Double.MAX_VALUE;
for (int i = begin; i <= end ; i += step) {
if(func.function(i) <= min)
min = func.function(i);
}
return min;
}
}
If you try with step between (0,1) this will be casted to int when adding to i, as a result you will add 0 to i in every iteration which will lead to infinite loop!
It will be infinite, because if step is 0 then adding zero to i would do nothing. No incrementation means infinite loop.
If step is 0 you're adding 0 to i meaning there's nothing being added to the value, and it gets stuck as infinite.
Since you supply step as a double, having values like 0.0001 is possible. However, since i is an integer the conversion is made from double to int, and 0.0001 -> 0. When you cast a value that's less than 1 and more than 0 as an integer, it becomes 0.
Change i to double and you can use decimal values to increment.
Alternatively, you could change step to int, but you wouldn't be able to use decimal steps in that case
A similar situation can be found with division or multiplication to demonstrate the double example:
(int) (500 * 0.0001) -> 0.05 as an integer = 0
So when you add a double step with a decimal value (exclusively a decimal value) to an integer, step becomes 0, 0 is added to i and it becomes infinite.
When it is (0,1), inside for loop you are verifying i <= end, this is all time passing condition. So, your loop won't break.
