840 Pro and HyperX 3K Pass 2 Petabytes With No Problem

[cesrai]

facepalm* KiB is 1024 bytes. A KB is 1000 bytes. A Megabyte is 1000 KB. Here's the kicker. A MiB is 1024 KB, not KiB. that's why anything above a KiB is not a perfect power of 2 (bytes of data). It's a factor of 1000 * 2^10.

Your wrong a MiB is 1024 KiB and a MiB is a 1048.58 KB.

[patrickjp93]

Your wrong a MiB is 1024 KiB and a MiB is a 1048.58 KB.

No, I'm not. Those standards he posted have not been adhered to by ANY storage manufacturer. It's the perfect myth. If only I had my 400 GiB Hitachi drive still... Every Unix OS displayed the actual byte count as being 1024*10^6

[cesrai]

It's like the standards people tried to implement on all SQL distributions. No one actually follows them.

People follow a lot of people follow them.

[cesrai]

No, I'm not. Those standards he posted have not been adhered to by ANY storage manufacturer. It's the perfect myth. If only I had my 400 GiB Hitachi drive still... Every Unix OS displayed the actual byte count as being 1024*10^6

Then do the math and show genius convert a MiB to a KiB and a MiB to a KB, yeah ignore Linux and OSX because they are wrong and you are right yeah.

[patrickjp93]

People follow a lot of people follow them.

Find me one living example and I'll bow right down. Best of luck on that endeavor.

[patrickjp93]

Then do the math and show genius convert a MiB to a KiB and a MiB to a KB, yeah ignore Linux and OSX because they are wrong and you are right yeah.

1MiB = 1.024MB = 1024KB = 1000KiB

 

Also, I'm running Debian, so I'm on Linux.

[cesrai]

Find me one living example and I'll bow right down. Best of luck on that endeavor.

HDD manufactures and Ram manufactures.

[cesrai]

1MiB = 1.024MB = 1024KB = 1000KiB

Also, I'm running Debian, so I'm on Linux.

You are wrong very wrong.

[cesrai]

1MiB = 1.024MB = 1024KB = 1000KiB

Also, I'm running Debian, so I'm on Linux.

My point still stands.

[patrickjp93]

HDD manufactures and Ram manufactures.

Neither of them do. Everyone switched over to the GB/TB system (no little i in the middle) system only a long time ago.

[patrickjp93]

My point still stands.

You have no proof, so your point doesn't stand at all.

[cesrai]

Neither of them do. Everyone switched over to the GB/TB system (no little i in the middle) system only a long time ago.

Really take ram stick out and read the label if you a 4GB stick you actually have 4GiB.

[patrickjp93]

Really take ram stick out and read the label if you a 4GB stick you actually have 4GiB.

No, I've actually tested that. You have 4GB. If you have ECC RAM, you get a few extra bits, but it's still not a perfect conversion.

[cesrai]

You have no proof, so your point doesn't stand at all.

I told you so the Math and you didn't and Linus and OSX calculate storage in decimal not binary.

[patrickjp93]

I told you so the Math and you didn't and Linus and OSX calculate storage in decimal not binary.

I've done the math and learned it from industry experts. You people quoted deprecated standards no one has ever followed fully in the first place.

[cesrai]

No, I've actually tested that. You have 4GB. If you have ECC RAM, you get a few extra bits, but it's still not a perfect conversion.

How did you test?

[cesrai]

I've done the math and learned it from industry experts. You people quoted deprecated standards no one has ever followed fully in the first place.

Where show me where did you do the math ?

[patrickjp93]

How did you test?

Turned off paging, looked at exactly how many bytes were in use before I launched a program which did nothing but allocate dynamic byte arrays of length 1 (char[] letter = new char[1]; in C++ without calling a delete statement, and there's no garbage collector either, so it's just a 1-byte memory leak for every time the program is executed). I launched it in a script which opened 1024 at a time. I watched the RAM fill up KiB by KiB, and the math doesn't work to suggest that it's a full GiB. It's plain GB.

 

 

Where show me where did you do the math ?

Go up 6 posts. You quoted where I did it. The actual and the standard do not match up.

[cesrai]

Turned off paging, looked at exactly how many bytes were in use before I launched a program which did nothing but allocate dynamic byte arrays of length 1. I launched it in a script which opened 1024 at a time. I watched the RAM fill up KiB by KiB, and the math doesn't work to suggest that it's a full GiB. It's plain GB.

Go up 6 posts. You quoted where I did it. The actual and the standard do not match up.

You just said ...=.....=.... You did no math at all you didn't use power you did nothing, Windows calculates in GiB and show in GB so if it was really 4GB the ram you have would be 3.... GB not 4GB.

[patrickjp93]

You just said ...=.....=.... You did no math at all you didn't use power you did nothing, Windows calculates in GiB and show in GB so if it was really 4GB the ram you have would be 3.... GB not 4GB.

OMG are we in first grade? I have to do everything? Fine.

 

1 KB = 1000 = 10^3 bytes, and 1 KiB = 1024 = 2^10 bytes. This is not in dispute. What is in dispute is what happens beyond that. 1MB = 1000KB = 10^6 bytes. 1MiB = 1024KB (NOT KiB) = (2^10)*1000= (2^10) * 10^3 bytes. That's it fully fleshed out. Happy now?

 

Also, Windows doesn't do it perfectly either, and it rounds up when looking at RAM. I can give you the source code for my test and tell you exactly how to duplicate the results. You have no clue what Windows is actually doing. Dump this in a notepad or whatever your favorite text editor is. 

int main(){

    char[] one = new char[1];

    return 0;

}

 

Disable all paging in your OS (the system where RAM deposits current system data to the disk drive to save space when stored memory has not been used recently).

 

Now, create either a Python or Perl or Lisp script that runs this program 1024 times (a KiB of allocated memory per run of the script).

 

Now take note of how much your ram usage is exactly by using the appropriate Windows commandline commands are to view current system status(I use Linux whenever I do command line anything so figure that out yourself). Watch your RAM run out at a rate faster than it should if you really had x * (2^10)^3 bytes in it.

 

Alternatively, you can take in text by having the arguments int argc, char*[] argv in your main header, and pull in a number for each execution from the command line. It's your choice.

[Shyxlol]

It's like the standards people tried to implement on all SQL distributions. No one actually follows them.

 

Ok, so you're saying that "by defininition" it technically is/should be 2^10 => 2^20 => 2^30 => etc. but in reality was/is (?) used as 2^10 => 10^3 * 2^10 => 10^6 * 2^10 => etc.?

I don't have means/time to actually test anything so I'll just say that a defined prefix like Ki/Mi/Gi which is defined as the "binary prefix" (from what I found) should still work like this: Mi= Ki*Ki, Gi= Mi*Ki etc. Just like the sheet/my table would suggest. But according to you is/has been used differently? Am I correct with that assumption?

[patrickjp93]

Ok, so you're saying that "by defininition" it technically is/should be 2^10 => 2^20 => 2^30 => etc. but in reality was/is (?) used as 2^10 => 10^3 * 2^10 => 10^6 * 2^10 => etc.?

I don't have means/time to actually test anything so I'll just say that a defined prefix like Ki/Mi/Gi which is defined as the "binary prefix" (from what I found) should still work like this: Mi= Ki*Ki, Gi= Mi*Ki etc. Just like the sheet/my table would suggest. But according to you is/has been used differently? Am I correct with that assumption?

Yes. No one in the memory or storage industries follows the true binary form.

[cesrai]

1KB=1000=10^3 bytes and 1 KiB=1024=2^10 bytes I agree on previous, 1MB=1000KB=10^6 bytes and 1MiB=1024KiB=20^2 bytes not 2^10*1000 bytes, Do you get it now are you happy? 

@patrickjp93

[cesrai]

Ok, so you're saying that "by defininition" it technically is/should be 2^10 => 2^20 => 2^30 => etc. but in reality was/is (?) used as 2^10 => 10^3 * 2^10 => 10^6 * 2^10 => etc.?

I don't have means/time to actually test anything so I'll just say that a defined prefix like Ki/Mi/Gi which is defined as the "binary prefix" (from what I found) should still work like this: Mi= Ki*Ki, Gi= Mi*Ki etc. Just like the sheet/my table would suggest. But according to you is/has been used differently? Am I correct with that assumption?

He is wrong read above ^^^ Sorry Assumed you were talking about space and what is 10^3 * 2^10 ??

Edited December 7, 2014 by cesrai

[Shyxlol]

He is wrong read above ^^^

 

Well, he is saying that according to the definition it technically is true but the industry uses it differently. I have neither evidence to prove nor disprove that so I'll just stick with that statement.