如何计算 CPU 缓存的标记、索引字段的大小（以位为单位）？

发布于 2024-12-17 01:25:18 字数 500 浏览 5 评论 0原文

我正在编写一个 CPU 缓存模拟器，它将获取以字节为单位的缓存大小、每个缓存行的长度（以字节为单位）以及组数/groups 在缓存中。

我已经写了大部分内容，但几个小时以来我一直在努力解决的是计算出需要左/右移动多少位才能提取标签和索引给定地址的字段。

例如，给定地址48，我需要确定标签和索引。

这是我提取标签的方法，但我很确定它是不正确的。

int extractTag(int address, int sets){

    int bits = exp2(sets); // number of bits to shift: 2^sets

    unsigned int tag;
    int tag = address >> (32 - bits);    

    return tag;
}

原文

I'm writing a CPU cache emulator that will take the size of the cache in bytes, the length of each cache line in bytes, and the number of sets/groups in the cache.

I have most of it written, but what I've been struggling with for hours is to figure out how many bits I need to shift left/right to extract the tag and index fields of the given address.

For example, given the address 48, I need to determine the tag and index.

Here's what I have for extracting the tag, but I'm pretty sure it's incorrect.

int extractTag(int address, int sets){

    int bits = exp2(sets); // number of bits to shift: 2^sets

    unsigned int tag;
    int tag = address >> (32 - bits);    

    return tag;
}

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梦屿孤独相伴 2024-12-24 01:25:18

假设您有 SETS 组 BLOCK_SIZE 行。地址可以拆分为 tag:index:offset，其中 log2(BLOCK_SIZE) 位用于偏移量，log2(SETS) 位用于索引，其余位用于标记。

您可以这样计算 log2：

int ilog2 (int x)
{
    int result = 0;

    while (x != 0) {
        result++;
        x = x >> 1;
    }
    return result;
}

因此您最终得到：

int extract_tag (int address, int sets, int block_size)
{
    int offset_bits = ilog2(block_size);
    int index_bits = ilog2(sets);

    int tag = address >> (index_bits + offset_bits);
    return tag;
}

int extract_index (int address, int sets, int block_size)
{
    int offset_bits = ilog2(block_size);
    int index_bits = ilog2(sets);

    int index = (address >> offset_bits) & ((1 << index_bits) - 1);
    return index;
}

int extract_offset (int address, int sets, int block_size)
{
    int offset_bits = ilog2(block_size);

    int offset = address & ((1 << offset_bits) - 1);
    return offset;
}

Let's say you have SETS groups of BLOCK_SIZE lines. An address can be split in tag:index:offset with log2(BLOCK_SIZE) bits for the offset, log2(SETS) for the index and the rest for the tag.

You can calculate log2 like this:

int ilog2 (int x)
{
    int result = 0;

    while (x != 0) {
        result++;
        x = x >> 1;
    }
    return result;
}

Thus you end up with:

int extract_tag (int address, int sets, int block_size)
{
    int offset_bits = ilog2(block_size);
    int index_bits = ilog2(sets);

    int tag = address >> (index_bits + offset_bits);
    return tag;
}

int extract_index (int address, int sets, int block_size)
{
    int offset_bits = ilog2(block_size);
    int index_bits = ilog2(sets);

    int index = (address >> offset_bits) & ((1 << index_bits) - 1);
    return index;
}

int extract_offset (int address, int sets, int block_size)
{
    int offset_bits = ilog2(block_size);

    int offset = address & ((1 << offset_bits) - 1);
    return offset;
}

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