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I read the code of pieh-ejdsch and everything has been optimized to perform fewer sets and normalized to be more streamlined. Even operations have been cleverly wrapped with 0x, 0x100 prefixes, etc. But it performs the same number of sets regardless of the string length. One or two sets should be el...

In the meantime, to understand what pieh-ejdsch did (by the way, you did a great job!), I tested the functions. I'm attaching the code for testing. pieh-ejdsch, can you list the tricks you used and their contribution to speed? I optimized for the lower part of the length because it's more likely to ...

(Only yesterday I deleted most of my test codes to make room on my desktop again :lol: No problem as I published it here as well...) Hmm, I can't see any noticeable differences yet. Not even if I raise the number of iterations from 1,000 to 10,000. Perhaps If I increase them even more? The updated ...

I add this variant made by T3rror, because it introduces intelligent ideas without losing speed. Indeed it should be faster because it supposes the non -existence of double quotes around the variables Hi Einst - I was looking over the final StrLen macro version you posted on dostips and noticed an e...

Thanks for the detailed and understandable explanation of escaping double quotes.

The rewritten code with spaces is a really good idea; I'll keep it in mind.

Francesco

Yes, what Antonio mentioned was used in that thread testing the performance of the set command, and in some cases there was a difference. In the meantime, I added the limit to 8191. It lost a bit in the lower range, but I don't understand why; further testing is needed. There was also a bug with the...

The last one I made isn't the fastest, but a compressed version that uses the previous fast one. Here's the fast one (it took me a few days to create the macro on one line like you did; I'm going crazy.) :: make sure we are in the spotlight: :: run as administrator for realtime priority!!! :: /affin...

even shorter. setlocal set "args=%1 %2" for /f "tokens=1-3" %%1 in ("!args! FEDCBA9876543210") do ( if defined %%~1 ( if "" neq "!%%~1:~255!" ( if "" neq "!%%~1:~4095!" ( set "$=!%%~1:~4095!" ) else set "$=!%%~1!" set "$scale=!$:~255,1!!$:~511,1!!$:~767,1!!$:~1023,1!!$:~1279,1!!$:~1535,1!!$:~1791,1!...

I was referring to the slowing down of the "set": Why does SET performance degrade as environment size grows? This is a version that reduces the macro size while still maintaining almost all of its speed. It's a good fit for me. setlocal set "args=%1 %2" for /f "tokens=1,2" %%1 in ("!args!") do ( if...

I tried to optimize the upper part even more, this is the code and the times setlocal set "args=str len" for /f "tokens=1,2" %%1 in ("!args!") do ( if defined %%~1 ( if "" neq "!%%~1:~255!" ( if "" neq "!%%~1:~4095!" ( set "$=!%%~1:~4095!" ) else set "$=!%%~1!" set "$scale=!$:~255,1!!$:~511,1!!$:~76...

I made another version that's even faster, but I discarded it because the macro was too long and the env for Win 10 is a problem.

I haven't tried it on Windows 11, but they told me the set problem related to the env size is no longer there.

Do you know if this is the case?

I took the last one you made and optimized it a bit. I removed a "Set" from everything and removed another set for some lengths. Also in the upper part from 4096 to 8192. I compared it, and it's faster than the last one. Of course, I compared the code without the macros. I think you'll have to figur...

It's been a year since the version I had on my PC. Yours seems faster. I haven't had time to study it. I added it to the old tests and I think the added checks and setlocal slow it down a bit. Now I'm studying it to see if there's anything I can improve on in terms of speed. @echo off setlocal Enabl...

... "_SIN=a-a*a/1920*a/312500+a*a/1920*a/15625*a/15625*a/2560000-a*a/1875*a/15360*a/15625*a/15625*a/16000*a/44800000" this seemed to have an algorithm protection ex) 1+3 =4 or (1/2 + 3/40*20)*200/100=4 there is no protection, if you calculate the third term for x=7000, with your formula it gives 13...

@miskox I'm trying that formula 10000*y/10/2 and mathematically since there is a multiplication by 10000 (which has 2 and 5 among the prime factors) the division by these should give an exact integer result without remainder or fractional part. That is, for any Y in that formula there is no remainde...