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Forked from jboner/latency.txt
Created August 22, 2019 02:29
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Latency Numbers Every Programmer Should Know
Latency Comparison Numbers (~2012)
----------------------------------
L1 cache reference 0.5 ns
Branch mispredict 5 ns
L2 cache reference 7 ns 14x L1 cache
Mutex lock/unlock 25 ns
Main memory reference 100 ns 20x L2 cache, 200x L1 cache
Compress 1K bytes with Zippy 3,000 ns 3 us
Send 1K bytes over 1 Gbps network 10,000 ns 10 us
Read 4K randomly from SSD* 150,000 ns 150 us ~1GB/sec SSD
Read 1 MB sequentially from memory 250,000 ns 250 us
Round trip within same datacenter 500,000 ns 500 us
Read 1 MB sequentially from SSD* 1,000,000 ns 1,000 us 1 ms ~1GB/sec SSD, 4X memory
Disk seek 10,000,000 ns 10,000 us 10 ms 20x datacenter roundtrip
Read 1 MB sequentially from disk 20,000,000 ns 20,000 us 20 ms 80x memory, 20X SSD
Send packet CA->Netherlands->CA 150,000,000 ns 150,000 us 150 ms
Notes
-----
1 ns = 10^-9 seconds
1 us = 10^-6 seconds = 1,000 ns
1 ms = 10^-3 seconds = 1,000 us = 1,000,000 ns
Credit
------
By Jeff Dean: http://research.google.com/people/jeff/
Originally by Peter Norvig: http://norvig.com/21-days.html#answers
Contributions
-------------
'Humanized' comparison: https://gist.github.com/hellerbarde/2843375
Visual comparison chart: http://i.imgur.com/k0t1e.png
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#include <stdlib.h>
#include <stdio.h>

// Number to guess: How big of an array (in bytes)
// can we allocate and fill with 5s in a second?
// The catch: We do it out of order instead of in order.
int main(int argc, char **argv) {
    int NUMBER, i;
    NUMBER = atoi(argv[1]);

    char* array = malloc(NUMBER);
    int j = 1;
    for (i = 0; i < NUMBER; ++i) {
        j = j * 2;
        if (j > NUMBER) {
            j = j - NUMBER;
        }
        // array[j] = j; // 没有局部访问性
        array[i] = j; // 有空间局部访问性有效使用缓存
    }

    printf("%d", array[NUMBER / 7]);
    // so that -O2 doesn't optimize out the loop

    return 0;
}

Answer:108-109
主存访问是100ns级别,L2缓存是10ns级别,L1缓存是0.5ns级别

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