Help with bitwise operations

Question

I want to doublecheck some of my logic against a 3rd party function that I am using and I'm not sure if I've got the bitwise logic figured out correctly or not. Can someone give me a range of values for the variable 'intValue' in each scenario that will cause each conditional to return true? thanks!

        if ((intValue < 0 && ((intValue & 0xFFFFFF80) == 0xFFFFFF80)) ||
            (intValue & 0x0000007F) == intValue) {

        }
        else if ((intValue < 0 && ((intValue & 0xFFFF8000) == 0xFFFF8000)) ||
            (intValue & 0x00007FFF) == intValue) {

        }
        else if ((intValue < 0 && ((intValue & 0xFF800000) == 0xFF800000)) ||
            (intValue & 0x007FFFFF) == intValue) {

        }
        else {

        }

How about you tell us what *you* think they are, and we'll go from there! — Oliver Charlesworth, Jul 08 '11 at 21:49

score 2 · Accepted Answer · answered Jul 08 '11 at 23:58

2

if (intValue == (SByte)intValue) {
     // matches -128...127, i.e. values that fit in one byte
}
else if (intValue == (Int16)intValue) {
     // matches -32768..32767, values that fit in two bytes
}
else if (intValue == ((intValue << 8) >> 8)) {
     // matches -2**23..(2**23-1), values that fit in three bytes
}
else {
     // matches -2**31..(2**31-1), values that need four bytes
}

Please note that all the intValue < 0 tests are completely redundant if the variable type is signed.

answered Jul 08 '11 at 23:58

Ben Voigt

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I think you mean "if the variable type is _un_ signed"? – Lii Feb 18 '15 at 09:47
1

@Lii: No, I didn't. The subsequent test looks at the sign bit. – Ben Voigt Feb 18 '15 at 14:22
And I guess if the variable is unsigned the tests are equally redundant since it can't hold negative values? – Lii Feb 18 '15 at 14:30
1

@Lii: Also correct. But redundant in a different way (implied by the data type, not the other clause in the test). – Ben Voigt Feb 18 '15 at 14:32

score -1 · Answer 2 · edited May 23 '17 at 10:34

-1

Here's a hint:

int intValue = int.Parse("FFFFFF80", System.Globalization.NumberStyles.HexNumber);

See:

C# convert integer to hex and back again

edited May 23 '17 at 10:34

Community

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answered Jul 08 '11 at 23:40

Chuck Savage

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Help with bitwise operations

2 Answers2