CODEWITHSUNDEEP
cbit-manipulation
Francis
Francis

Reputation: 343

Changing only some bits with mask

I need help with finding the correct bit mask. 0xbc61 has to be 0xbcaa, but I am not sure how to create the bit mask to affect only the part I have to change. I also tried splitting the value but wasn't sure how to "glue" them with the new one...

unsigned short int left, right;
unsigned short int temp = 0xbc61;
unsigned short int newValue = 0xaa;
right= temp & 0xff; // split
left= (temp >> 8) // split

// temp needs to be 0xbcaa
// and I also need it the other way round, so 0xbc61 -> 0xaa61

Upvotes: 0

Views: 1561

Answers (3)

0___________
0___________

Reputation: 68004

to change any arbitraty bits to anothers:

uint32_t changebits(uint32_t val, uint32_t mask, uint32_t newvalue)
{
    val &= ~mask;
    val |= (newvalue & mask);
    return val
}

 changebits(0xe323da98, 0b00000110011000010100000100000000, 0xde45ab67);

it will set only the bits given by the second parameters to the same bits from the third parameter.

in your case 

changebits(0xbc61, 0xff, 0xaa);

Upvotes: 1

Alex Lop.
Alex Lop.

Reputation: 6875

You can use the XOR here:

0x61 is 0b01100001

0xaa is 0b10101010

To convert one value into another just use XOR with 0xCB (0b11001011):

unsigned short mask = 0xbc61;
....
// convert to new mask 0xbcaa
mask ^= 0xCB; // mask now is 0xbcaa
...
// convert back to 0xbc61
mask ^= 0xCB; // mask now is 0xbc61

NOTE: A XOR 1 = !A and A XOR 0 = A

Here is the demo: https://ideone.com/UIxPDV

Upvotes: 1

Bartek Lew
Bartek Lew

Reputation: 101

You mean newValue = (oldValue & 0xff00) | 0xaa;? Or you need to use minimal mask? If so, write your number in binary and check it. :)

Upvotes: 0

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