I have the following code...
int Val=-32768; String Hex=Integer.toHexString(Val);
This equates to ffff8000
int FirstAttempt=Integer.parseInt(Hex,16); // Error "Invalid Int" int SecondAttempt=Integer.decode("0x"+Hex); // Error "Invalid Int"
So, initially, it converts the value -32768 into a hex string ffff8000, but then it can't convert the hex string back into an Integer.
In .Net
it works as I'd expect, and returns -32768
.
I know that I could write my own little method to convert this myself, but I'm just wondering if I'm missing something, or if this is genuinely a bug?
Since double has longer range than int data type, java automatically converts int value to double when the int value is assigned to double. 1. Java implicit conversion from int to double without typecasting.
toHexString() is a built-in function in Java which returns a string representation of the integer argument as an unsigned integer in base 16. The function accepts a single parameter as an argument in Integer data-type.
just use constructor of Number class.
int val = -32768; String hex = Integer.toHexString(val); int parsedResult = (int) Long.parseLong(hex, 16); System.out.println(parsedResult);
That's how you can do it.
The reason why it doesn't work your way: Integer.parseInt
takes a signed int, while toHexString
produces an unsigned result. So if you insert something higher than 0x7FFFFFF
, an error will be thrown automatically. If you parse it as long
instead, it will still be signed. But when you cast it back to int, it will overflow to the correct value.
It overflows, because the number is negative.
Try this and it will work:
int n = (int) Long.parseLong("ffff8000", 16);
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