CODEWITHSUNDEEP
cbit-manipulation
A. Vieira
A. Vieira

Reputation: 1281

Change 4 middle bits of a byte in C

I'm trying to change the 4 middle bits of a byte to correspond to the High nibble of another byte:

Suppose we start with:

In = 0bABCDEFGH
Out = 0bXXXXXXXX // Some random byte

I want:

Out = 0bXXABCDXX

Leaving whatever other bits were in Out's extremes unchanged.

How can I do this?

Note: The 'X' represents any bit, 0 or 1, just to distinguish what came from the input.

I got to:

(0b00111100 & (IN>>2)) = 0b00ABCD00

, which filters the high nibble and centers it but then what? How can I move it to Out?

Upvotes: 4

Views: 1311

Answers (3)

user5550963
user5550963

Reputation:

simple:

out &= 0b11000011;
out |= (in >> 2 & 0b00111100);

out &= 0b11000011 sets out to 0bxx0000xx preserving 2 most significant bits and 2 least significant bits. in >> 2 shifts input by 2 giving us 0xYYABCDEF, YY could be 00 or 11 depending on what A is. To get rid of YY and EF we do & 0b00111100.

As pointed by @JB 0B is not standard notation, thus you should use something else, most preferably hex 0x notation. See this for more info.

Thus using hex this would be:

out &= 0xC3;
out |= (in >> 2 & 0x3C)

here is conversion table

`0xf` is `0b1111`
`0x3` is `0b0011`
`0xc` is `0b1100`

Upvotes: 7

John Bollinger
John Bollinger

Reputation: 181724

There are multiple alternatives. From a high-level perspective, you could

  • force the four middle bits of Out off, prepare a mask from In as show in your question, and combine Out and mask via bitwise OR (|)
  • force the four middle bits of Out off, prepare a mask from In as show in your question, and combine Out and mask via bitwise EXCLUSIVE OR (^)
  • force the four middle bits of Out on, prepare a mask from In similarly to how you do now, but with the outer bits on, and combine Out and mask via bitwise AND (&)
  • use a series of shifts, masks, and addition or bitwise OR operations to build up the wanted result section by section

Forcing bits off is achieved by bitwise AND with a mask that has 0s at (only) the positions you want to turn off.

Forcing bits on is achieved by bitwise OR with a mask that has 1s at (only) the positions you want to turn on.

You already seem to have a handle on shifting, though you do need to be careful there if you happen to be shifting objects of signed types. Prefer to use unsigned types for bit manipulation wherever possible.

Upvotes: 1

Acorn
Acorn

Reputation: 26194

Assuming in and out are unsigned char, and that CHAR_BIT == 8:

out = (out & 0xC3) | ((in >> 2) & 0x3C);

i.e. 4 operations in total.

Upvotes: 4

Related Questions