5 ms·
Mathematically wrong, unless restated correctly: "accessing ALL memory at a given cache level is O(N^[1/3])". After restatement: trite and useless. Accessing a
by dmitrygr 1y ago
Mathematically wrong, unless restated correctly: "accessing ALL memory at a given cache level is O(N^[1/3])". After restatement: trite and useless.
Accessing any given one byte in the address space is amortized to O(1)
- deleted 1y ago[deleted]
- curiouscoding 1y agoTry measuring it! You'll quickly see that the latency of addressing n bytes of memory is definitely not O(1). See eg "The Myth of RAM": https://www.ilikebigbits.com/2014_04_21_myth_of_ram_1.html https://www.ilikebigbits.com/2014_04_21_myth_of_ram_1.html
- dmitrygr 1y ago100% wrong 8^(1/3) is 2 So according to OP (and you by agreement), upgrading my machine from 4GB of RAM to 32GB should double my RAM access time. Obviously, and demonstrably, nothing of the sort occurs!
- tabender 1y agoBut for another example, we often say array access is O(1) - but if you increased the size of your array by 8x (assuming you were already operating at the speed of light) you would see that increase!
- dmitrygr 1y agoWhat nonsense are you going on about? My Dram controller has no idea as to which bytes are part of my array and which bytes aren’t, so even if it tried to play along with your claims, it wouldn’t know how big my array was declared an thus which accesses and by how much to slow down.