what does a[0] = addr & 0xff?

Question

i'm currently learning from the book "the shellcoder's handbook", I have a strong understanding of c but recently I came across a piece of code that I can't grasp.

Here is the piece of code:

          char a[4];
          unsigned int addr = 0x0806d3b0;
          a[0] = addr & 0xff;
          a[1] = (addr & 0xff00) >> 8; 
          a[2] = (addr & 0xff0000) >> 16;
          a[3] = (addr) >> 24;

So the question is what does this, what is addr & 0xff (and the three lines below it) and what makes >> 8 to it (I know that it divides it 8 times by 2)? Ps: don't hesitate to tell me if you have ideas for the tags that I should use.

Answer 1

The variable addr is 32 bits of data, while each element in the array a is 8 bits. What the code does is copy the 32 bits of addr into the array a, one byte at a time.

Lets take this line:

a[1] = (addr & 0xff00) >> 8; 

And then do it step by step.

1. addr & 0xff00 This gets the bits 8 to 15 of the value in addr, the result after the operation is 0x0000d300.
2. >> 8 This shifts the bits to the right, so 0x0000d300 becomes 0x000000d3.
3. Assign the resulting value of the mask and shift to a[1].