What is actually happening with the XOR mask?

Your mask is your test case of which you will compare the number against. XOR means one or the other is true but not both. Let’s take a basic example with the intention of flipping the bits. I am going to leave out the “0b” part and just use 0’s and 1’s.

Mask: 111 Num: 110

From RIGHT to LEFT using XOR:

1 and 0 produce a 1 because only 1 of them is a 1. 1 and 1 produce a 0 because both are 1 1 and 1 produce a 0 because both are 1

Result: 001 –> This is exactly the opposite as what we started with which was 110.

Why make the mask all 1’s? Let’s try this again but we’ll make the mask all 0’s.

Mask: 000 Num: 110

From RIGHT to LEFT using XOR:

0 and 0 produce a 0 because both are 0 0 and 1 produce a 1 because only 1 of them is a 1 0 and 1 produce a 1 because only 1 of them is a 1

Result: 110 –> We started with 110 so the bits did not change.

You can try this on paper with other mask combinations but only using XOR with a mask that has the bits turned on will result in the flipping of bits.

You can use the “~” operator but XOR lets you manage individual bits which has many more applications.

First,, you have the desired number number = 0b01110101. Using a mask, you can flip bits using XOR. If, for example, you wanted to change the 3rd and 6th bit of the number, then mask = 0b00100100 0b01110101 ^ 0b00100100 = 0b01010001 By doing XOR on the number and the mask, you get the answer

0 ^ 0 = 0
1 ^ 0 = 1
1 ^ 1 = 0
1 ^ 0 = 1
0 ^ 0 = 0
1 ^ 1 = 0
0 ^ 0 = 0
1 ^ 0 = 1

Definition of XOR: If there is 0 ^ 0, then the result is 0, just like OR. if there is 0 ^ 1 or 1 ^ 0, then the result is 1, just like OR. However, if there is 1 ^ 1, the result is 0. You can use not, but you can’t modify individual bits.

a = 0b0001 mask = 0b1111 a^mask=0b1110 flipped see?