Question about casting

[PaulStat]

Just playing around with casting to see the effect it has if you don't do it properly!

For example 123456789^2 = 15241578750190521

The following code produces the result -1757895751

public class Order {
	public int int1 = 123456789;
	public int int2 = 123456789;
	public long long1 = 0;
	public void calculateTotal(){
[B]		long long1 = (long)(int1*int2);[/B]
		System.out.println("Result: " + long1);
	}
}

Where as this produces the correct result

public class Order {
	public int int1 = 123456789;
	public int int2 = 123456789;
	public long long1 = 0;
	public void calculateTotal(){
[B]		long long1 = (long)int1*int2;[/B]
		System.out.println("Result: " + long1);
	}
}

Question is why does the first one produce an incorrect result, I would have thought casting the result of (int1*int2) would produce exactly the same thing as (long)int1*int2??????

 

[JIMA]

Hi,

In the first example the (int * int) will result in an integer being produced. The resulting number will be too large to hold in an int so the error will appear (turned to a negative number by the left-most bit being set). Casting to a long will have no effect on the outcome (it's the erroneous number produced ny the calculation that's being cast).

In the second case the first int is cast to a long so the datatypes in the multiplication are "long * int" which will result in a long being produced. As the long can hold the number produced without overflowing the correct answer will be seen.

I think that's about right.....

Jim

 

[PaulStat]

Hmm yeah that makes sense, I did come up with a similar answer myself about half an hour after posting (all apart from "turned to a negative number by the left-most bit being set").

So the following lines would also produce the same [correct] result?

[B]long long1 = int1*(long)int2;[/B]

[B]long long1 = (long)int1*(long)int2;[/B]

Actually, CAN you cast the second variable or would that produce a syntax error, hmmmm

 

[|Ric|]

They should both give the correct result (i.e. no overflow) and both are fine. You can cast any primitive where ever it is in an expression