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Bitshifting of a number in C


By : Rick Carmichael
Date : September 12 2020, 05:00 AM
Does that help The long in the C program's memory will be raw binary, so the first step is to split it up into decimal format digits. This is done by repeatedly dividing the number by 10 and grabbing the remainder. The LS digit of the number is always n % 10, and when you do n / 10 you "shift" the decimal number one digit to the right. Once you have an array of digits, printing it as BCD is trivial - it's just printing a binary number in string format.
Start by coding the above program and once you have it working, try to achieve the same with whatever artificial nonsense requirements you have from your teacher.
code :


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What does this c# bitshifting method do?


By : Karim Essam
Date : March 29 2020, 07:55 AM
Does that help Extract from the ID the bit at the position passed.
Position should be 1-8
code :
ID = 128;  // 10000000
getBit(8); // returns 1

ID = 127;  // 01111111
getBit(8); // returns 0

C#: looping over an xor key and bitshifting?


By : Fohte
Date : March 29 2020, 07:55 AM
like below fixes the issue I have a key stored as a 4-byte integer. Say , First convert the key into a byte array:
code :
byte[] keyBytes;
unchecked
{
    keyBytes = new byte[]{(byte)(key >> 0),
                          (byte)(key >> 8),
                          (byte)(key >> 16),
                          (byte)(key >> 24);
}
for (int i = 0; i < barray.length; i++)
{
    barray[i] ^= keyBytes[i%4];
}

BitShifting for uint in c#


By : immograf
Date : March 29 2020, 07:55 AM
I hope this helps . It would only throw that exception if in a checked context. See: http://msdn.microsoft.com/en-us/library/y3d0kef1(v=vs.80).aspx.
No exception:
code :
uint u1 = uint.MaxValue - 1000;
byte[] b = BitConverter.GetBytes(u1);
uint u2 = BitConverter.ToUInt32(b, 0);
uint u3 = (uint) (b[0] | (b[1] << 8) | (b[2] << 16) | (b[3] << 24));
checked
{
    uint u1 = uint.MaxValue - 1000;
    byte[] b = BitConverter.GetBytes(u1);
    uint u2 = BitConverter.ToUInt32(b, 0);
    uint u3 = (uint) (b[0] | (b[1] << 8) | (b[2] << 16) | (b[3] << 24));
}
checked
{
    unchecked
    {
        uint u1 = uint.MaxValue - 1000;

        byte[] b = BitConverter.GetBytes(u1);
        uint u2 = BitConverter.ToUInt32(b, 0);
        uint u3 = (uint) (b[0] | (b[1] << 8) | (b[2] << 16) | (b[3] << 24));

        Console.WriteLine(u1 + " " + u2 + " " + u3);
    }
}

negative number when bitshifting


By : wasiur rahman
Date : March 29 2020, 07:55 AM
I wish this help you First of all, turn on Option Strict(see also: what's an option strict and explicit?), so you have to define a type for everything. It'll be worth it in the long run.
Now that we have to create types for everything, your function can look like this:
code :
Public Function MergeABGR(A as Byte, R as Byte, G as Byte, B as Byte) as UInt32
    Return (Convert.ToUInt32(A) << 24) + (Convert.ToUInt32(R) << 16) + (Convert.ToUInt32(G) << 8) + B
End Function

Why first bitshifting left and then right, instead of AND-ing?


By : Ram
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , Assuming that the size of int is 32 bits, then there is no need to use shifts. Indeed, bitwise & with a mask would be more readable, more portable and safer.
It should be noted that left-shifting on negative signed integers invokes undefined behavior, and that left-shifting things into the sign bits of a signed integer could also invoke undefined behavior. C11 6.5.7 (emphasis mine):
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