[Java] Loss of Precision

Posted in xn--kfs74mzzid01b.com edit by wktd on January 7th, 2009

I got these errors in my function that prints out a byte in 1's and 0's. I don't get why it's saying that there will be a loss of precision if a byte type in Java is 8 bits and I assigned exactly 8 bits to that byte type.

==============================================

dcllnx17> javac PrintSimilarity.java
./Neighbors.java:25: possible loss of precision
found : int
required: byte
byte ander = 0x80;
^
./Neighbors.java:30: possible loss of precision
found : int
required: byte
ander = 0x80 >>> bit;
^
2 errors

// Name: charToBin
// Input: a single byte
// Output: none
// Purpose: prints out the character input in binary
public static void charToBin(byte theByte)
{
// ander = 1000 0000
byte ander = 0x80;

// tests each bit of input by ANDing 1 with each bit
for(int bit = 0; bit < 8; bit++)
{
ander = 0x80 >>> bit;
if((ander & theByte) == ander)
System.out.print("1");
else
System.out.print("0");
}
}

try

ander = (byte)(0x80 >>> bit);

You have to keep in mind that 0x80 is just another way of specifying an integer. In this case it fits in 1 byte, but to Java it's still an integer. You're trying to assign an integer (4 bytes) to a variable of the type byte (1 byte), so Java is warning you for possible loss of precision. So what you need to do is cast your integer to a byte, explicitly telling Java that you're ok with the "loss" of precision:

byte ander = (byte) 0x80;The same thing happens on this line:
ander = 0x80 >>> bit;You're performing a bitshift on 0x80 (an integer) which results in another integer which you are then again assigning to a byte. So here, as eirche pointed out before, you have to cast it to a byte as well.

That fixes the 2nd error, but the first error still remains.

byte ander = 0x80;

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[Java] Loss of Precision

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