Given a page size of 512 bytes

Given a page size of 512 bytes, what is the page number and the offset for the virtual address 0x1234?

The correct answer and explanation is:

To determine the page number and offset for the virtual address 0x1234, given a page size of 512 bytes, follow these steps:

Step 1: Convert the address to binary

The virtual address 0x1234 in hexadecimal is equivalent to 0001 0010 0011 0100 in binary.

Step 2: Understand the structure of the address

Since the page size is 512 bytes, we need to determine how many bits are required to represent the offset within a page.

The page size of 512 bytes is equivalent to 292^9 bytes, meaning 9 bits are used for the offset within a page. The remaining bits in the address represent the page number.

Step 3: Divide the address into the page number and the offset

Given that the page size is 512 bytes, the least significant 9 bits of the address represent the offset within the page. The remaining higher-order bits represent the page number.

The binary address 0001 0010 0011 0100 is split as follows:

  • The rightmost 9 bits represent the offset.
  • The leftmost bits represent the page number.

So, the binary address 0001 0010 0011 0100 (0x1234) can be split into:

  • Offset: The rightmost 9 bits are 011 0100, which is 0x34 in hexadecimal (52 in decimal).
  • Page number: The remaining bits are 0001 0010, which is 0x12 in hexadecimal (18 in decimal).

Step 4: Conclusion

  • The page number is 0x12 (or 18 in decimal).
  • The offset is 0x34 (or 52 in decimal).
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