Why are page sizes always powers of 2?

[zak100]

Hi,

I was going through a solution and I found the above question with the answer:

Quote

Recall that paging is implemented by breaking up an address into a page
and offset number. It is most efficient to break the address into X page
bits and Y offset bits, rather than perform arithmetic on the address
to calculate the page number and offset. Because each bit position
represents a power of 2, splitting an address between bits results in
a page size that is a power of 2.

Is there a better answer for this? What I understand is that we are using a binary number system, that's why page sizes should be powers of 2. Somebody please guide me whether I am right or wrong?

Zulfi.

[Sensei]

Multiplication or division by 2^x can be replaced by bitwise left shift or bitwise right shift.

https://en.wikipedia.org/wiki/Bitwise_operation

They are extremely fast, in comparison to regular multiplication or division by arbitrary integer (not to mention floating point).

 

 

 

[zak100]

Hi,

Thanks a lot.

 

Zulfi.

[Strange]

 

16 minutes ago, zak100 said:

Is there a better answer for this? What I understand is that we are using a binary number system, that's why page sizes should be powers of 2.

Exactly. 

If you have a 32 bit address space, for example, then the total number of bytes you can address is 2³².

If you use, say, 5 bits for the address within a page (the offset) and the remaining 27 bits for the page address, then you will have 2²⁷ pages each of 2⁵ bytes. 

[Sensei]

In binary numeral system, to clear the least, or the most significant bits, there can be used bitwise AND operator e.g. in C/C++:

a = x & ~( 1 << y - 1 );

and to get reminder there can be used yet another bitwise AND operator e.g.

b = x & ( 1 << y - 1 );

 

In the normal circumstances you (or CPU) would have to do modulo e.g.:

a = ( x / y ) * y;

and e.g.

b = x - a * y;

Multiplication and division by arbitrary integer are slower than bitwise operators.

 

[Sensei]

2 hours ago, Sensei said:

and e.g.

b = x - a * y;

Ah, a is already premultiplied by y..

so, correctly I should write:

b = x - a;