I have a problem with C’s bitwise operations and can’t “merge” two numbers together.
Suppose I have the number 0x0A (or any 2 hexadecimal digits) and I want to replace the left digit with 0xF for example, That is:
00001010 -> 11111010 in terms of bitwise.

What I tried to do is the following:

1. We drop the left digit of the 2-digit hex number by left shifting (sizeof(int)*8-4) times, thus leaving the rightmost bits that we care about, and then right shifting (unsigned) so that we get only the right hex digit in the end. In the example above it won’t matter as it’ll yield ‘A’ which is the same as 0x0A.
2. The second step is to “merge” the digit we got in step one and the shifted hex digit we want to use, and we’ll do that by using an OR gate.

Thus:

Hex number before and after step one: A (1010)
Desired hex digit after shifting: F0 (1111 0000)
OR gate:

00001010
OR 11110000
= 11111010 = 0xFA

However, for some reason, my C code does all of the above correctly, and just when applying OR it gives the result 0xFFFFFFFA (8-byte).

Is there something that I’m missing on how C works with bitwise?

Note that I use sizeof(int) because the code should be able to account for 4 or 8 byte int.

Thanks for help in advance!

int setHexDigit(char* digitPairPtr, int digitPos, char hexDigit)
{
    int return_value = 0;
    if( (hexDigit>='0' && hexDigit<='9') || (hexDigit>='A' && hexDigit <='F'))
    {
        if(digitPos==0)
        {   
            hexDigit = ((hexDigit<='9' && hexDigit>='0') ? (hexDigit-'0') : ( (hexDigit>='A' && hexDigit<='F') ? (hexDigit-('A'-10)) : (-1))); //THIS LINE IS NOT IMPORTANT 
            *digitPairPtr = ((unsigned)*digitPairPtr<<((sizeof(int))*8-4))>>((sizeof(int))*8-4); //Drop left digit.
            hexDigit = (hexDigit << 4);
            *digitPairPtr = *digitPairPtr | hexDigit;
            return_value=1;
        }
        else
        {
            //For later
        }
    }
    return return_value;
}

int main()
{
    char c = 0x0A;
    setHexDigit(&c,0,'F');
    printf("\n%X", c);
    return 0;
}