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Conversion from int to 4 byte unsigned int

I'm trying to convert an int to a 4 byte unsigned ints. I was reading a post and I'm not sure what comparison does. I know that its a sub mask, but I'm not sure when to use the & and when to use the |. Lets use the number 6 as an example, if the LSB is the number 6. Why would we do 

6 & 0xFF// I know that we are comparing it to 11111111

when do we use the OR operator? I'm still not sure how to use the & nor |

Solution

  • x & 0xFF will set all bits of x to zero except for the last byte (which stays the same). If you had used a bitwise or (|), it would leave the bits of x set, and set all the bits of the last byte to 1.

    Typically, the comparison will be something like (x & 0xFF) == x. This is to make sure that the first three bytes of x are all 0.