Let's take a sample write packet : Suppose that the CPU wrote the value
0x12345678
to the physical address0xfdaff040
using 32-bit addressingThis example is from this site (I didn't understand the explanations in the original post)
[31 : 2]
An address of an aligned, 32-bit chunk always has two zero bits at the end of the address. You can think of this as either writing the address of the chunk to the 32-bit slot or as writing the addresses divided by four to bits 2 through 31 of the address. The result is the same either way since dividing by four is equivalent to shifting two bit positions to the right.