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Java Forum / General / February 2007Comparing byte values doesn't work
Hi, I have a doubt.Why doesn't the following loop work? Logically,the if statement should have been satisfied atleast once. class Test { public static void main(String[] args) { for (byte b = Byte.MIN_VALUE; b < Byte.MAX_VALUE; b++) { if (b == 0x90) System.out.print("Printed"); }//loop-ends } } It works with: if(b==(byte)0x90) But, 0x90 = 00....0000 1001 0000 +144 (byte)0x90 = 1001 0000 -0x70 -0x70 is well within the byte range.Then what makes it unequal to all byte values? Thanks in advance. > -0x70 is well within the byte range.Then what makes it unequal to > all byte values? The operation is *integer* comparison, so you are comparing each byte with 0x90 (144), not -0x70. As you've already discovered, you need to cast 0x90 to byte for the comparison to succeed. Read about "binary numeric promotion" here: http://java.sun.com/docs/books/jls/third_edition/html/expressions.html#5198 http://java.sun.com/docs/books/jls/third_edition/html/conversions.html#170983 /gordon  Signature[ don't email me support questions or followups ] g o r d o n + n e w s @ b a l d e r 1 3 . s e > The operation is *integer* comparison, so you are comparing each byte > with 0x90 (144), not -0x70. > Read about "binary numeric promotion" here:http://java.sun.com/docs/books/jls/third_edition/html/expressions.htm...http://j ava.sun.com/docs/books/jls/third_edition/html/conversions.htm... On Feb 14, 11:55 pm, Eric Sosman <Eric.Sos...@sun.com> wrote: > (byte)0x90 == -0x70 == -112 is within the range of byte > values. (byte)0x90 != 0x90; the first is negative and > the second is positive. Thanks, the language specifications helped a lot. On Feb 15, 12:41 am, "Daniel Pitts" <googlegrou...@coloraura.com> wrote: > Like others have said, the comparison first converts to integers. > 0x90 = 192, your loop ranges from -128 to 126. [quoted text clipped - 5 lines] > } > } Is the casting of the first operand used to force byte calculations? I tried println(b) with both ways and got the same result -112. class Test { public static void main(String[] args) { for (int b = Byte.MIN_VALUE; b <= Byte.MAX_VALUE; b++) { if (b == (byte)0x90) { System.out.println("Printed"); System.out.println(b); } }//loop-ends } } > Is the casting of the first operand used to force byte calculations? No, the comparison is done using ints. Follow the links I posted. Byte operands are promoted to int before the comparison is made. > I tried println(b) with both ways and got the same result -112. What both ways? I think you'll find that these give different results: System.out.println(0x90); System.out.println((byte)0x90); Casting with (byte)b where b is already a byte doesn't change anything. /gordon Signature[ don't email me support questions or followups ] g o r d o n + n e w s @ b a l d e r 1 3 . s e > > Is the casting of the first operand used to force byte calculations? > [quoted text clipped - 13 lines] > Casting with (byte)b where b is already a byte doesn't change > anything. Yes, i went through the link and soon did conclude that my earlier loop was undergoing a byte to integer conversion. But the next post i was referring to another snip werein it used if ((byte)b == (byte)0x90) Thanks again. On Feb 15, 12:41 am, "Daniel Pitts" <googlegrou...@coloraura.com> wrote: > Like others have said, the comparison first converts to integers. > 0x90 = 192, your loop ranges from -128 to 126. > Two things to do: > for (int b = Byte.MIN_VALUE; b <= Byte.MAX_VALUE; b++) { > if ((byte)b == (byte)0x90) { > System.out.println("Printed"); > } > } > > > Is the casting of the first operand used to force byte calculations? [...] > if ((byte)b == (byte)0x90) I'm not sure if I've understood what you are asking, but I hope this helps... What that code will do is 1) Cast b to a byte. Since b is already a byte, this has absolutely no effect and will be ignored by the compiler. 2) Cast 0x90 to a byte, resulting in the decimal byte value -112. (Actually that will be done at compile-time rather than at runtime, but it makes no difference here). 3) Convert /both/ byte values to 32-bit integers. 4) Compare those integers. There is no way that you can compare two bytes directly in Java -- they are /always/ promoted to integers before the comparison. (And before all other arithmetic operations, such as subtraction.) (BTW, if all this seems somewhat overcomplicated to you, then don't worry -- I think it's overcomplicated too. The real problem is that in Java (against all good sense) the 'byte' datatype is signed.) -- chris On Feb 15, 6:55 am, "Chris Uppal" <chris.up...@metagnostic.REMOVE- THIS.org> wrote: > > > > Is the casting of the first operand used to force byte calculations? > [...] [quoted text clipped - 24 lines] > > -- chris Actually, you missed (twice) his reference to MY post, in which b is an int. rajeev.nospam@gmail.com wrote On 02/14/07 13:38,: > Hi, > I have a doubt.Why doesn't the following loop work? [quoted text clipped - 18 lines] > -0x70 is well within the byte range.Then what makes it unequal to all > byte values? The range of a byte goes from -128 == -0x80 through 127 == 0x7F. 0x90 == 144 is outside that range: it is a larger value than any byte can hold, so no byte can ever be equal to it and the test will never be satisfied. (byte)0x90 == -0x70 == -112 is within the range of byte values. (byte)0x90 != 0x90; the first is negative and the second is positive. Besides which, your loop leaves one value untested. SignatureEric.Sosman@sun.com On Feb 14, 10:38 am, "rajeev.nos...@gmail.com" <rajeev.nos...@gmail.com> wrote: > Hi, > I have a doubt.Why doesn't the following loop work? [quoted text clipped - 21 lines] > > Thanks in advance. Like others have said, the comparison first converts to integers. 0x90 = 192, your loop ranges from -128 to 126. Two things to do: for (int b = Byte.MIN_VALUE; b <= Byte.MAX_VALUE; b++) { if ((byte)b == (byte)0x90) { System.out.println("Printed"); } } Free MagazinesGet these publications absolutely FREE for up to 12 months. There are no hidden fees and no obligation. 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