gpt4 book ai didi

operating-system - 缓存中 "block size"的概念

转载 作者:行者123 更新时间:2023-12-02 09:01:00 26 4
gpt4 key购买 nike

我刚刚开始学习直接映射和集合关联缓存的概念。我有一些非常基本的疑问。开始吧。

假设地址长 32 位,并且我有一个 32KB 缓存、64 字节 block 大小和 512 帧,那么“ block ”内实际存储了多少数据?如果我有一条从内存位置加载值的指令,并且该值是 16 位整数,那么 64 字节 block 之一现在仅存储 16 位(2 字节)整数值。该 block 内的其他 62 个字节又如何呢?如果我现在有另一个加载指令也加载一个 16 位整数值,则该值现在根据加载地址进入另一个帧的另一个 block (如果该地址映射到前一个指令的同一帧,则前一个值将被逐出)并且该 block 再次仅存储 64 字节中的 2 字节)。正确的?

如果这看起来是一个非常愚蠢的疑问,请原谅我,这只是我想正确地理解我的概念。

最佳答案

我输入这封电子邮件是为了让某人解释缓存,但我认为您可能会发现它也很有用。

You have 32-bit addresses that can refer to bytes in RAM. You want to be able to cache the data that you access, to use them later.

Let's say you want a 1-MiB (220 bytes) cache.

What do you do?

You have 2 restrictions you need to meet:

  1. Caching should be as uniform as possible across all addresses. i.e. you don't want to bias toward any particular kind of address.
    • How do you do this? Use remainder! With mod, you can evenly distribute any integer over whatever range you want.
  2. You want to help minimize bookkeeping costs. That means e.g. if you're caching in blocks of 1 byte, you don't want to store 4 bytes of data just to keep track of where 1 byte belongs to.
    • How do you do that? You store blocks that are bigger than just 1 byte.

Let's say you choose 16-byte (24-byte) blocks. That means you can cache 220 / 24 = 216 = 65,536 blocks of data.

You now have a few options:

  • You can design the cache so that data from any memory block could be stored in any of the cache blocks. This would be called a fully-associative cache.
  • The benefit is that it's the "fairest" kind of cache: all blocks are treated completely equally.
  • The tradeoff is speed: To find where to put the memory block, you have to search every cache block for a free space. This is really slow.
  • You can design the cache so that data from any memory block could only be stored in a single cache block. This would be called a direct-mapped cache.
  • The benefit is that it's the fastest kind of cache: you do only 1 check to see if the item is in the cache or not.
  • The tradeoff is that, now, if you happen to have a bad memory access pattern, you can have 2 blocks kicking each other out successively, with unused blocks still remaining in the cache.
  • You can do a mixture of both: map a single memory block into multiple blocks. This is what real processors do -- they have N-way set associative caches.

Direct-mapped cache:

Now you have 65,536 blocks of data, each block being of 16 bytes.
You store it as 65,536 "rows" inside your cache, with each "row" consisting of the data itself, along with the metadata (regarding where the block belongs, whether it's valid, whether it's been written to, etc.).

Question: How does each block in memory get mapped to each block in the cache?

Answer: Well, you're using a direct-mapped cache, using mod. That means addresses 0 to 15 will be mapped to block 0 in the cache; 16-31 get mapped to block 2, etc... and it wraps around as you reach the 1-MiB mark.

So, given memory address M, how do you find the row number N? Easy: N = M % 220 / 24.
But that only tells you where to store the data, not how to retrieve it. Once you've stored it, and try to access it again, you have to know which 1-MB portion of memory was stored here, right?

So that's one piece of metadata: the tag bits. If it's in row N, all you need to know is what the quotient was, during the mod operation. Which, for a 32-bit address, is 12 bits big (since the remainder is 20 bits).

So your tag becomes 12 bits long -- specifically, the topmost 12 bits of any memory address.
And you already knew that the lowermost 4 bits are used for the offset within a block (since memory is byte-addressed, and a block is 16 bytes).
That leaves 16 bits for the "index" bits of a memory address, which can be used to find which row the address belongs to. (It's just a division + remainder operation, but in binary.)

You also need other bits: e.g. you need to know whether a block is in fact valid or not, because when the CPU is turned on, it contains invalid data. So you add 1 bit of metadata: the Valid bit.

There's other bits you'll learn about, used for optimization, synchronization, etc... but these are the basic ones. :)

关于operating-system - 缓存中 "block size"的概念,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/8107965/

26 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com