缓存命中矩阵乘法

Question

https://youtu.be/o7h_sYMk_oc?t=1963

在这个视频中，他解释说检索远处的数据会导致更差的缓存行利用率，然后他接着说了一条我不明白的行。 “所以处理器将引入 64 个字节来对特定数据进行操作。然后它会忽略该缓存行上 8 个浮点字中的 7 个并转到下一个” 他这是什么意思。

Answer 1

缓存通常基于缓存行。当数据读入缓存时，是通过读取一个完整的缓存行来完成的。因此，如果缓存行包含 64 个字节，则处理器 HW 确保将 64 个连续字节从内存读取到缓存中。如果一个浮点双精度数是 8 个字节，那么单个缓存行可以容纳 8 个双精度数。

现在，如果您的代码使用连续的双打，缓存访问将是：

Access double located in Addr    --> Miss, 64 bytes read into the cache (slow)
Access double located in Addr+1  --> Hit (fast)
Access double located in Addr+2  --> Hit (fast)
Access double located in Addr+3  --> Hit (fast)
Access double located in Addr+4  --> Hit (fast)
Access double located in Addr+5  --> Hit (fast)
Access double located in Addr+6  --> Hit (fast)
Access double located in Addr+7  --> Hit (fast)
Access double located in Addr+8  --> Miss, 64 bytes read into the cache (slow)
Access double located in Addr+9  --> Hit (fast)
Access double located in Addr+10 --> Hit (fast)
Access double located in Addr+11 --> Hit (fast)
Access double located in Addr+12 --> Hit (fast)
Access double located in Addr+13 --> Hit (fast)
Access double located in Addr+14 --> Hit (fast)
Access double located in Addr+15 --> Hit (fast)
Access double located in Addr+16 --> Miss, 64 bytes read into the cache (slow)
Access double located in Addr+17 --> Hit (fast)
...

所以在这里你有 1 次慢速读取，然后是 7 次快速读取，因为你的程序使用连续的双精度数。

但是，如果您的程序总是使用彼此间隔 8 个双精度数（又名 64 字节）的双精度数，则您的模式将为：

Access double located in Addr    --> Miss, 64 bytes read into the cache (slow)
Access double located in Addr+8  --> Miss, 64 bytes read into the cache (slow)
Access double located in Addr+16 --> Miss, 64 bytes read into the cache (slow)
...

在这里你只会得到缓慢的读取，你不会从缓存系统中获得任何好处。

缓存命中矩阵乘法

Cache hits on matrix multiplication

c

memory

performance