import
   math, strformat, sequtils, times

const
   B* = 8
   Ki* = 2 ^ 10
   Mi* = Ki * Ki

type
   CacheInfo* = object
      size*, lineSize*: int


template incEvery*[T](arr: var openArray[T]; n, step: int) =
   for i in countup(0, n * step - 1, step):
      inc arr[i]

template time*(s: untyped): auto =
   let t0 = cpuTime()
   s
   cpuTime() - t0

template repeat*(count: int; s: untyped): auto =
   for i in 1 .. count:
      s

const
   cacheInfos* = [CacheInfo(size: 32 * Ki, lineSize: 64),
      CacheInfo(size: 256 * Ki, lineSize: 64), CacheInfo(size: 2 * Mi, lineSize: 64)]

var bytes: array[4 * Mi, byte]

proc test(ci: CacheInfo; step, n: int) =
   let repeatCount = 50000
   let t = time(repeat(repeatCount, bytes.incEvery(n, step)))
   let T = t / n.float * 1000.0
   let b = n.float * repeatCount.float / t
   echo fmt"step={step:2}, n={n:5}: T={T:6.4f}ms, b={b/Mi.float:6.2f}MiB/s"

proc main() =
   block:
      let ci = cacheInfos[0]
      test(ci, 1, 2 * ci.size)
   for i, ci in cacheInfos:
      echo fmt"L{i + 1}: size={ci.size div Ki}KiB, line size={ci.lineSize}B"
      test(ci, ci.lineSize, 2 * ci.size div ci.lineSize)