# New solutions to the old problem

• Tutorial

##### 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.
Entrance:
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.
Output:
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.

Algorithm:
1) Go from older notebooks to younger ones
2) Take a notebook and convert it to a byte with an extension of zero.
4) If the result is more than 39h then
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
std
lodsb
mov bl,al
and al,0fh
call digit
cld
stosb
mov al,bl
shr al,4
call digit
ret
digit:
cmp al,39h
jnb _zero-nine
;digit greater than 9
_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
``````mov cx,8
mov si,value
mov di,hexstr
add si,cx ;highest byte of value
dec si
std
lodsb
mov bl,al
and al,0fh
call digit
cld
stosb
mov al,bl
shr al,4
call digit
ret
digit:
mov ah,30h
aaa
ret
``````

Another way is to use the XLATB command.
Decision No. 3
Xlatb
``````         mov cx,8
mov si,value
mov di,hexstr
mov	bx,hextable
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.
``````         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
``````

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.
Unpuck
``````         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.
Carry
``````         mov rdx,0f0f0f0f0f0f0f0fh
mov rcx,0606060606060606h
mov rbx,rax
and rax,rdx
mov rdi,rax ;копия пригодится позже
shr rbx,4
and rbx,rdx
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
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
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.
Deinterleaving
``````        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
psrlq	xmm1,4
pand	xmm1,[efes]
pxor	xmm9,xmm9
psubb	xmm9,xmm1
pand	xmm9,[sevens]
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.
CPU12345678
AMD C-60290195290120105120140290

1. Lodsb stosb with jnb
2. Lodsb stosb with xlatb
3. General purpose registers with shrd
4. General purpose registers with punpck
5. SSE 64 bit
6. SSE 64 bit unaligned
7. SSE 128 bit
8. 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
movdqa	xmm3,xmm2
psrlq   xmm1,4
pand    xmm1,[efes]
psrlq	xmm3,4
pxor    xmm9,xmm9
pand	xmm3,[efes]
psubb   xmm9,xmm1
pxor	xmm10,xmm10
pand    xmm9,[sevens]
psubb	xmm10,xmm3
pand	xmm10,[sevens]
movdqa  [hexstr],xmm0
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
``````