Masking a bit in C [duplicate] - c

Question

I am fairly new to C programming, and I encountered bit masking. What is the general concept and function of bit masking?
Examples are much appreciated.

Answer 1

A mask defines which bits you want to keep, and which bits you want to clear.
Masking is the act of applying a mask to a value. This is accomplished by doing:
Bitwise ANDing in order to extract a subset of the bits in the value
Bitwise ORing in order to set a subset of the bits in the value
Bitwise XORing in order to toggle a subset of the bits in the value
Below is an example of extracting a subset of the bits in the value:
Mask: 00001111b
Value: 01010101b
Applying the mask to the value means that we want to clear the first (higher) 4 bits, and keep the last (lower) 4 bits. Thus we have extracted the lower 4 bits. The result is:
Mask: 00001111b
Value: 01010101b
Result: 00000101b
Masking is implemented using AND, so in C we get:
uint8_t stuff(...) {
uint8_t mask = 0x0f; // 00001111b
uint8_t value = 0x55; // 01010101b
return mask & value;
}
Here is a fairly common use-case: Extracting individual bytes from a larger word. We define the high-order bits in the word as the first byte. We use two operators for this, &, and >> (shift right). This is how we can extract the four bytes from a 32-bit integer:
void more_stuff(uint32_t value) { // Example value: 0x01020304
uint32_t byte1 = (value >> 24); // 0x01020304 >> 24 is 0x01 so
// no masking is necessary
uint32_t byte2 = (value >> 16) & 0xff; // 0x01020304 >> 16 is 0x0102 so
// we must mask to get 0x02
uint32_t byte3 = (value >> 8) & 0xff; // 0x01020304 >> 8 is 0x010203 so
// we must mask to get 0x03
uint32_t byte4 = value & 0xff; // here we only mask, no shifting
// is necessary
...
}
Notice that you could switch the order of the operators above, you could first do the mask, then the shift. The results are the same, but now you would have to use a different mask:
uint32_t byte3 = (value & 0xff00) >> 8;

Answer 2

Masking means to keep, change, or remove a desired part of information. Let’s see an image-masking operation; like this masking operation is removing anything that is not skin:
We are doing an AND operation in this example. There are also other masking operators—OR and XOR.
Bitmasking means imposing mask over bits. Here is a bitmasking with AND—
1 1 1 0 1 1 0 1 input
(&) 0 0 1 1 1 1 0 0 mask
------------------------------
0 0 1 0 1 1 0 0 output
So, only the middle four bits (as these bits are 1 in this mask) remain.
Let’s see this with XOR—
1 1 1 0 1 1 0 1 input
(^) 0 0 1 1 1 1 0 0 mask
------------------------------
1 1 0 1 0 0 0 1 output
Now, the middle four bits are flipped (1 became 0, 0 became 1).
So, using a bitmask, we can access individual bits (examples). Sometimes, this technique may also be used for improving performance. Take this for example-
bool isOdd(int i) {
return i%2;
}
This function tells if an integer is odd/even. We can achieve the same result with more efficiency using a bit-mask—
bool isOdd(int i) {
return i&1;
}
Short Explanation: If the least significant bit of a binary number is 1 then it is odd; for 0 it will be even. So, by doing AND with 1 we are removing all other bits except for the least significant bit, i.e.:
55 -> 0 0 1 1 0 1 1 1 input
(&) 1 -> 0 0 0 0 0 0 0 1 mask
---------------------------------------
1 <- 0 0 0 0 0 0 0 1 output