New solutions to the old problem
Or transfer the bike to jet propulsion
There is one very old task whose age is equal to the age of the American Standard Code for the Exchange of Information. More specifically, it is the task of converting an integer to its hexadecimal representation of an ASCII string.
In this publication, we will consider the conversion of an unsigned sixty-four-bit integer into a string of fixed length without truncating the leading zeros.
The task at first glance seems elementary. It would be so if the ASCII table was different. But we have what we have.
All solutions will only be for IA-32 and Intel 64 architecture.
Consider the input-output data and the algorithm for solving this problem.
A 64-bit unsigned integer that occupies 8 bytes at addresses from low to high. The digits of a number are arranged in the same order. Each bit takes 4 bits, each byte has two.
A string of ASCII characters, each occupying one byte. Each byte represents one digit of the original number. The order of the characters is the first byte (lowest address) - the most significant bit of the original number.
1) Go from older notebooks to younger ones
2) Take a notebook and convert it to a byte with an extension of zero.
3) Add 30h
4) If the result is more than 39h then
4.1) Add another 17 (decimal)
4.2) Go to 5
4.3) Otherwise go to 5
5) Save the received byte to the string
6) While not all notebooks have been processed, go to 2
Solution No. 1
mov cx,8 mov si,value mov di,hexstr add si,cx ;highest byte of value dec si next_tetrade: std lodsb mov bl,al and al,0fh call digit cld stosb mov al,bl shr al,4 call digit loop next_tetrade ret digit: add al,30h cmp al,39h jnb _zero-nine ;digit greater than 9 add al,11h _zero-nine: ret
Like most decisions in the forehead, it is not distinguished by either beauty or speed. Home ugliness is a conditional transition that bypasses just one command. There are two ways to bring beauty.
The first is to use ancient “magic” as a sequence of AAA AAD 17 commands.
Solution # 2
AAA AAD 17
mov cx,8 mov si,value mov di,hexstr add si,cx ;highest byte of value dec si next_tetrade: std lodsb mov bl,al and al,0fh call digit cld stosb mov al,bl shr al,4 call digit loop next_tetrade ret digit: mov ah,30h aaa aad 11h ret
Another way is to use the XLATB command.
Decision No. 3
mov cx,8 mov si,value mov di,hexstr mov bx,hextable add si,cx dec si m3: std lodsb mov ah,al and al,0fh xlatb cld stosb mov al,ah shr al,4 xlatb stosb loop m3 ret hextable db "0123456789ABCDEF"
Both solutions are almost identical. But Solution No. 2 will not work in 64-bit processor mode, due to the elimination of support for AAA and AAD commands in it.
But is it really possible to process 8 bytes at a time, will we accept processing only 4 bits at a time?
Are there any ways to turn 9 (1001) into 39h (00111001) and A (1010) into 41h (01000001)?
Let's try to reveal the essence of a pair of AAA AAD teams, and pick up a replacement for them.
AAA AAD Replacement
mov bl,0ah xor ah,ah div bl ; al - частное, ah -остаток. Понятно, что для цифр больше 9 частное будет равно 1. mov bh,al ; shl al,4 ; и из этой единицы мы формируем 11h add al,bh add al,30h ; voila!
This code, of course, is not suitable for our purposes, but it provides valuable information. It shows that you can get a transfer unit for numbers greater than 9. And it shows how to use this unit later. Now, if it were possible to turn each notebook into a byte, then these bytes could be processed in parallel. Eight digits at a time!
Let rax contain the original number. To isolate the younger notebooks, you just need to perform conjunction with the mask. And so as not to lose the older notebooks, copy them to rbx.
mov rdx,0f0f0f0f0f0f0f0fh mov rbx,rax and rax,rdx shr rbx,4 and rbx,rdx
Unfortunately, it will not work to obtain the desired transfer by division. Are there any other methods?
Actually, elementary: 10h-0ah = 6. It is enough to add 6 to each byte and we will get the necessary unit in the senior notebook.
mov rdx,0f0f0f0f0f0f0f0fh mov rcx,0606060606060606h mov rbx,rax and rax,rdx mov rdi,rax ;копия пригодится позже shr rbx,4 and rbx,rdx add rax,rcx shr rax,4 and rax,rdx ; теперь в тех байтах, которые соответствуют цифрам больше 9, находятся единицы. В других - нули.
Unlike the previous method of obtaining units of transfer using division, instead of the remainder of division, we still have the original figure. That is, where the number A was - there it will remain, and not turn into 0. Therefore, you need to add not 11h, but 41h-3ah = 7.
And now the next task arises, how to make seven out of one? Yes, so as not to affect neighboring bytes. After all, 7 = 0111b, which means you can get what you need with two left shifts and two disjunctions.
One to seven
mov rsi,rax shl rsi,1 or rax,rsi ;11b shl rsi,1 or rax,rsi ; 111b
Now put the pieces together and see what happens
Unsigned to hex ver 0.1
mov rax,[value] mov rdx,0f0f0f0f0f0f0f0fh mov rcx,0606060606060606h mov rbx,rax mov rbp,3030303030303030h shr rbx,4 and rax,rdx and rbx,rdx mov rdi,rax mov r9,rbx add rax,rcx add rbx,rcx shr rax,4 shr rbx,4 and rax,rdx and rbx,rdx mov rsi,rax mov r8,rbx shl rsi,1 shl r8,1 or rax,rsi or rbx,r8 shl rsi,1 shl r8,1 or rax,rsi or rbx,r8 add rax,rbp add rbx,rbp add rax,rdi add rbx,r9 mov [hexstr],rax mov [hexstr+8],rbx
If you compile and run this code, one trouble will be revealed - the order of the numbers in the line is not in order. Firstly, the numbers go back and forth, and secondly, even and odd numbers are grouped together. You can, of course, give it to conclusion, and so, but we are honest and still put things in order.
mov rcx,4 highpart: rol rbx,8 shrd [hexstr],rbx,8 rol rax,8 shrd [hexstr],rax,8 loop highpart mov rcx,4 lowpart: rol rbx,8 shrd [hexstr+8],rbx,8 rol rax,8 shrd [hexstr+8],rax,8 loop lowpart ret
From what they wanted to leave, they came to that. Again byte processing in 64 mode. In order to flip the bytes, put them backwards, Intel has long made the bswap command. But for deinterlacing you have to look towards MMX, SSE and their development. And there is such a team and its name is punpcklbw. We use our finds.
Deinterleaving ver. 2
bswap rax bswap rbx mov [hexstr],rax mov [hexstr+8],rbx movdqu xmm0,[hexstr] movdqu xmm1,[hexstr+8] punpcklbw xmm1,xmm0 movdqu [hexstr],xmm1
Stop stop stop. If we started using SSE, maybe there is something else useful? It could be to rewrite our code entirely on SSE.
Unsigned to hex ver 1.1
movdqu xmm0,[value] pxor xmm1,xmm1 punpcklbw xmm0,xmm1 movdqa xmm1,xmm0 pand xmm1,[efes] psllq xmm0,4 pand xmm0,[efes] por xmm0,xmm1 movdqa xmm1,xmm0 paddb xmm1,[sixes] psrlq xmm1,4 pand xmm1,[efes] pxor xmm9,xmm9 psubb xmm9,xmm1 pand xmm9,[sevens] paddb xmm0,xmm9 paddb xmm0,[zeroes] movdqu [hexstr],xmm0 mov rax,[hexstr] mov rbx,[hexstr+8] bswap rax bswap rbx mov [hexstr],rbx mov [hexstr+8],rax ret efes: dq 0f0f0f0f0f0f0f0fh dq 0f0f0f0f0f0f0f0fh zeroes: dq 3030303030303030h dq 3030303030303030h sixes: dq 0606060606060606h dq 0606060606060606h sevens: dq 0707070707070707h dq 0707070707070707h
Here we simplified the unpacking, and organized the obtaining of the sevens in a different way, using one subtraction and one conjunction.
But what did we win at all? Compare the performance of each of the methods.
|Core 2 Quad Q8200||670||600||150||77||70||77||76||1170|
- Lodsb stosb with jnb
- Lodsb stosb with xlatb
- General purpose registers with shrd
- General purpose registers with punpck
- SSE 64 bit
- SSE 64 bit unaligned
- SSE 128 bit
- Lodsb stosb with xlatb 128 bit
Values in parrots are the average number of processor ticks.
If the numbers on Intel fit well in theory, then on AMD they are somewhat mysterious. A pleasant surprise was the high speed SSE code on the processor from Intel. You can safely increase the bit depth of the converted numbers to 256 bits with a slight increase in the required time, since there are still many free xmm registers in 64-bit mode. In general, initially it would seem a sequential task, it became possible to solve a very parallel method.
The inverse problem of converting a hexadecimal string to a number is solved no less entertainingly.
For a snack:
SSE 128 bit
movdqa xmm0,[value] pxor xmm1,xmm1 movdqa xmm2,xmm0 punpcklbw xmm0,xmm1 movdqa xmm1,xmm0 punpckhbw xmm2,xmm1 pand xmm1,[efes] movdqa xmm3,xmm2 psllq xmm0,4 pand xmm3,[efes] pand xmm0,[efes] psllq xmm3,4 por xmm0,xmm1 pand xmm2,[efes] movdqa xmm1,xmm0 por xmm2,xmm3 paddb xmm1,[sixes] movdqa xmm3,xmm2 psrlq xmm1,4 paddb xmm3,[sixes] pand xmm1,[efes] psrlq xmm3,4 pxor xmm9,xmm9 pand xmm3,[efes] psubb xmm9,xmm1 pxor xmm10,xmm10 pand xmm9,[sevens] psubb xmm10,xmm3 paddb xmm0,xmm9 pand xmm10,[sevens] paddb xmm0,[zeroes] paddb xmm2,xmm10 movdqa [hexstr],xmm0 paddb xmm2,[zeroes] mov rax,[hexstr] movdqa [hexstr+16],xmm2 mov rbx,[hexstr+8] mov rcx,[hexstr+16] bswap rax mov rdx,[hexstr+16+8] bswap rbx bswap rcx mov [hexstr],rbx bswap rdx mov [hexstr+8+16],rax mov [hexstr+8],rcx mov [hexstr+16],rdx ret