How to produce the cpu cache effect in C and java?

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情话喂你
情话喂你 2021-02-09 02:15

In Ulrich Drepper\'s paper What every programmer should know about memory, the 3rd part: CPU Caches, he shows a graph that shows the relationship between \"working set\" size an

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  •  野的像风
    2021-02-09 03:12

    This answer isn't an answer, but more of a set of notes.

    First, the CPU tends to operate on cache lines, not on individual bytes/words/dwords. This means that if you sequentially read/write an array of integers then the first access to a cache line may cause a cache miss but subsequent accesses to different integers in that same cache line won't. For 64-byte cache lines and 4-byte integers this means that you'd only get a cache miss once for every 16 accesses; which will dilute the results.

    Second, the CPU has a "hardware pre-fetcher." If it detects that cache lines are being read sequentially, the hardware pre-fetcher will automatically pre-fetch cache lines it predicts will be needed next (in an attempt to fetch them into cache before they're needed).

    Third, the CPU does other things (like "out of order execution") to hide fetch costs. The time difference (between cache hit and cache miss) that you can measure is the time that the CPU couldn't hide and not the total cost of the fetch.

    These 3 things combined mean that; for sequentially reading an array of integers, it's likely that the CPU pre-fetches the next cache line while you're doing 16 reads from the previous cache line; and any cache miss costs won't be noticeable and may be entirely hidden. To prevent this; you'd want to "randomly" access each cache line once, to maximise the performance difference measured between "working set fits in cache/s" and "working set doesn't fit in cache/s."

    Finally, there are other factors that may influence measurements. For example, for an OS that uses paging (e.g. Linux and almost all other modern OSs) there's a whole layer of caching above all this (TLBs/Translation Look-aside Buffers), and TLB misses once the working set gets beyond a certain size; which should be visible as a fourth "step" in the graph. There's also interference from the kernel (IRQs, page faults, task switches, multiple CPUs, etc); which might be visible as random static/error in the graph (unless tests are repeated often and outliers discarded). There are also artifacts of the cache design (cache associativity) that can reduce the effectiveness of the cache in ways that depend on the physical address/es allocated by the kernel; which might be seen as the "steps" in the graph shifting to different places.

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