Sort the characters of a string in ascending order using masm32 programming language [duplicate] - sorting

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NASM assembly Insertion Sort
(1 answer)
How to swap 2 bytes in x86 assembly correctly?
(3 answers)
How to take string input in Assembly
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Closed 21 days ago.
I am trying to make a program that accepts user input and sorts the characters of a string input in ascending order.

The sorting algorithm used can vary, but a common approach is to use a bubble sort algorithm.
.data
string db 'sortstring',0
sorted db 20 dup(0)
.code
main proc
mov edx, offset string
mov ecx, lengthof string
call sort_string
; ...
; Output sorted string here
; ...
ret
main endp
sort_string proc
push ebp
mov ebp, esp
; ...
; Sort the characters of the string here
; ...
pop ebp
ret
sort_string endp

Related

cmp instruction and negative numbers [duplicate]

This question already has answers here:
Difference between JA and JG in assembly
(2 answers)
Closed 10 months ago.
Here is a code to sort the given array in assembly language
.model small
.stack 100h
.data ;roll number 2435
data1 db 66h, 2, 045h, 4, 040h, 3, -025h, 5, -010h, 011h
swap db 0
.code
mov ax, #data
mov ds, ax
start:
mov swap, 0
mov bx, 0
loop1:
mov al, [bx+data1]
mov cl, [bx+data1+1]
cmp al, [bx+data1+1] ;here is the problem when compare 66h with -025h
jbe noswap
mov dl, [bx+data1+1]
mov [bx+data1+1],al
mov [bx+data1], dl
mov swap, 1
noswap:
add bx,1
cmp bx,9
jne loop1
cmp swap,1
je start
mov ah, 04ch
int 21h
it compares all elements of array to sort in ascending order, but when it compares 66h with -025h, it implies that 66h is smaller and -025h is bigger and does not swap, mov to no swap lable
i have debugged it and found that at backend -025h is being stored as DB. How can I properly sort the array with negative number
Use the signed condition code in the conditional branch: e.g. jle instead of the unsigned condition code jbe. See https://sandpile.org/x86/cc.htm . If you ask the processor to do unsigned condition then it will see -025h as 0xDB, which is 219, so larger than 66h/102.
The processor doesn't read data declarations, so it doesn't see the minus sign.  Since it never sees declarations, it doesn't matter to the processor if you put db -25h or db 219 or db 0xdb — these will all populate the data with the same bit pattern value.
In C, for example, we give types to the variables and then the compiler (using language rules) generates machine code that accesses the variable consistently, i.e. as the size it was declared, and also as to whether signed or unsigned.
In assembly language we don't have variable declarations with expressive type information.  So, we must do the job that compilers do: use the proper size and signed'ness in the machine code for every access to variables.

What is the fastest way to swap the bytes of an unaligned 64 bit value in memory?

I have a large number of 64 bit values in memory. Unfortunately they might not be aligned to 64 bit addresses. My goal is to change the endianess of all those values, i.e. swapping/reversing their bytes.
I know about the bswap instruction which swaps the bytes of a 32 or 64 bit register. But as it requires a register argument, I cannot pass it my memory address. Of course I can first load the memory into register, then swap, then write it back:
mov rax, qword [rsi]
bswap rax
mov qword [rsi], rax
But is that even correct, given that the address might be unaligned?
Another possibility is to do the swaps manually:
mov al, byte [rsi + 0]
mov bl, byte [rsi + 7]
mov byte [rsi + 0], bl
mov byte [rsi + 7], al
mov al, byte [rsi + 1]
mov bl, byte [rsi + 6]
mov byte [rsi + 1], bl
mov byte [rsi + 6], al
mov al, byte [rsi + 2]
mov bl, byte [rsi + 5]
mov byte [rsi + 2], bl
mov byte [rsi + 5], al
mov al, byte [rsi + 3]
mov bl, byte [rsi + 4]
mov byte [rsi + 3], bl
mov byte [rsi + 4], al
That's obviously a lot more instructions. But is it slower, too?
But all in all I'm still pretty inexperienced in x86-64, so I wonder: What is the fastest way to byte swap a 64 bit value in memory? Is one of the two options I described optimal? Or is there a completely different approach that is even faster?
PS: My real situation is a bit more complicated. I do have a large byte array, but it contains differently sized integers, all densely packed. Some other array tells me what size of integer to expect next. So this "description" could say "one 32 bit int, two 64 bit ints, one 16 bit int, then one 64 bit int again". I am just mentioning this here to tell you that (as far as I can tell), using SIMD instructions is not possible as I actually have to inspect the size of each integer before reading.
What is the fastest way to byte swap a 64 bit value in memory?
The mov/bswap/mov version and the movbe/mov are about the same on most Intel processors. Based on the µop count, it seems movbe decodes to mov + bswap, except on Atom. For Ryzen, movbe may be better. Manually swapping around bytes is much slower, except in certain edge cases where a large load/store is very slow, such as when it crosses a 4K boundary pre-Skylake.
pshufb is a reasonable option even to replace a single bswap, though that wastes half of the work the shuffle could do.
PS: My real situation is a bit more complicated. I do have a large byte array, but it contains differently sized integers, all densely packed.
In this general case, with sizes dynamically taken from an other data stream, a new big issue is branching on the size. Even in scalar code that can be avoided, by byte-reversing a 64bit block and shifting it right by 8 - size, then merging it with the un-reversed bytes, and advancing by size. That could be worked out, but it's a waste of time to try that, the SIMD version will be better.
A SIMD version could use pshufb and a table of a shuffle-masks indexed by a "size pattern", for example an 8-bit integer where every 2 bits indicates the size of an element. pshufb then reverses the elements that are wholly contained in the 16-byte window that it's looking at, and leave the rest alone (those unchanged bytes at the tail will be written back too, but that's OK). Then we advance by the number of bytes that was actually processed.
For maximum convenience, those size patterns (as well as corresponding byte-counts) should be supplied in such a way that the actual Endianness Flipper itself can consume exactly one of them per iteration, without anything heavy such as extracting a byte-unaligned sequence of 8 bits and determining dynamically how many bits to consume. That's also possible, but at a significantly higher cost. About 4x as slow in my test, limited by the loop-carried dependency through "extract 8 bits at current bit-index" through "find bit-index increment by table lookup" and then into the next iteration: about 16 cycles per iteration, though still in 60% of the time that equivalent scalar code took.
Using an unpacked (1 byte per size) representation would make the extraction easier (just an unaligned dword load), but requires packing the result to index the shuffle mask table with, for example with pext. That would be reasonable for Intel CPUs, but pext is extremely slow on AMD Ryzen. A alternative that is fine for both AMD and Intel would be to do the unaligned dword read, then extract the 8 interesting bits using a multiply/shift trick:
mov eax, [rdi]
imul eax, eax, 0x01041040
shr eax, 24
An extra trick that should be used, in the Convenient Input case at least (otherwise we're stuck with a 5x worse performance anyway and this trick won't be relevant), is reading the data for the next iteration before storing the result of the current iteration. Without that trick, the store will often "step on the toes" of the load of the next iteration (because we advance less than 16 bytes, so the load reads some of the bytes that the store left unchanged but had to write anyway), forcing a memory dependency between them which hold up the next iteration. The performance difference is large, about 3x.
Then the Endianness Flipper could look something like this:
void flipEndiannessSSSE3(char* buffer, size_t totalLength, uint8_t* sizePatterns, uint32_t* lengths, __m128i* masks)
{
size_t i = 0;
size_t j = 0;
__m128i data = _mm_loadu_si128((__m128i*)buffer);
while (i < totalLength) {
int sizepattern = sizePatterns[j];
__m128i permuted = _mm_shuffle_epi8(data, masks[sizepattern]);
size_t next_i = i + lengths[j++];
data = _mm_loadu_si128((__m128i*)&buffer[next_i]);
_mm_storeu_si128((__m128i*)&buffer[i], permuted);
i = next_i;
}
}
For example, Clang 10 with -O3 -march=haswell turns that into
test rsi, rsi
je .LBB0_3
vmovdqu xmm0, xmmword ptr [rdi]
xor r9d, r9d
xor r10d, r10d
.LBB0_2: # =>This Inner Loop Header: Depth=1
movzx eax, byte ptr [rdx + r10]
shl rax, 4
vpshufb xmm1, xmm0, xmmword ptr [r8 + rax]
mov eax, dword ptr [rcx + 4*r10]
inc r10
add rax, r9
vmovdqu xmm0, xmmword ptr [rdi + rax]
vmovdqu xmmword ptr [rdi + r9], xmm1
mov r9, rax
cmp rax, rsi
jb .LBB0_2
.LBB0_3:
ret
LLVM-MCA thinks that takes about 3.3 cycles per iteration, on my PC (4770K, tested with a uniform mix of 1, 2, 4 and 8 byte sized elements) it was a little slower, closer to 3.7 cycles per iteration, but that's still good: that's just under 1.2 cycles per element.

Counting elements "less than x" in an array

Let's say you want to find the first occurrence of a value1 in a sorted array. For small arrays (where things like binary search don't pay off), you can achieve this by simply counting the number of values less than that value: the result is the index you are after.
In x86 you can use adc (add with carry) for an efficient branch-free2 implementation of that approach (with the start pointer in rdi length in rsi and the value to search for in edx):
xor eax, eax
lea rdi, [rdi + rsi*4] ; pointer to end of array = base + length
neg rsi ; we loop from -length to zero
loop:
cmp [rdi + 4 * rsi], edx
adc rax, 0 ; only a single uop on Sandybridge-family even before BDW
inc rsi
jnz loop
The answer ends up in rax. If you unroll that (or if you have a fixed, known input size), only the cmp; adc pair of instructions get repeated, so the overhead approaches 2 simple instructions per comparison (and the sometimes fused load). Which Intel microarchitecture introduced the ADC reg,0 single-uop special case?
However, this only works for unsigned comparisons, where the carry flag holds the result of the comparison. Is there any equivalently efficient sequence for counting signed comparisons? Unfortunately, there doesn't seem to be an "add 1 if less than" instruction: adc, sbb and the carry flag are special in that respect.
I am interested in the general case where the elements have no specific order, and also in this case where the array is sorted in the case the sortedness assumption leads to a simpler or faster implementation.
1 Or, if the value doesn't exist, the first greater value. I.e., this is the so called "lower bound" search.
2 Branch free approaches necessarily do the same amount of work each time - in this case examining the entire array, so this approach only make sense when the arrays are small and so the cost of a branch misprediction is large relative to the total search time.
PCMPGT + PADDD or PSUBD is probably a really good idea for most CPUs, even for small sizes, maybe with a simple scalar cleanup. Or even just purely scalar, using movd loads, see below.
For scalar integer, avoiding XMM regs, use SETCC to create a 0/1 integer from any flag condition you want. xor-zero a tmp register (potentially outside the loop) and SETCC into the low 8 of that, if you want to use 32 or 64-bit ADD instructions instead of only 8-bit.
cmp/adc reg,0 is basically a peephole optimization for the below / carry-set condition. AFAIK, there is nothing as efficient for signed-compare conditions. At best 3 uops for cmp/setcc/add, vs. 2 for cmp/adc. So unrolling to hide loop overhead is even more important.
See the bottom section of What is the best way to set a register to zero in x86 assembly: xor, mov or and? for more details about how to zero-extend SETCC r/m8 efficiently but without causing partial-register stalls. And see Why doesn't GCC use partial registers? for a reminder of partial-register behaviour across uarches.
Yes, CF is special for a lot of things. It's the only condition flag that has set/clear/complement (stc/clc/cmc) instructions1. There's a reason that bt/bts/etc. instructions set CF, and that shift instructions shift into it. And yes, ADC/SBB can add/sub it directly into another register, unlike any other flag.
OF can be read similarly with ADOX (Intel since Broadwell, AMD since Ryzen), but that still doesn't help us because it's strictly OF, not the SF!=OF signed-less-than condition.
This is typical for most ISAs, not just x86. (AVR and some others can set/clear any condition flag because they have an instruction that takes an immediate bit-position in the status register. But they still only have ADC/SBB for directly adding the carry flag to an integer register.)
ARM 32-bit can do a predicated addlt r0, r0, #1 using any condition-code, including signed less-than, instead of an add-with-carry with immediate 0. ARM does have ADC-immediate which you could use for the C flag here, but not in Thumb mode (where it would be useful to avoid an IT instruction to predicate an ADD), so you'd need a zeroed register.
AArch64 can do a few predicated things, including increment with cinc with arbitrary condition predicates.
But x86 can't. We only have cmovcc and setcc to turn conditions other than CF==1 into integers. (Or with ADOX, for OF==1.)
Footnote 1: Some status flags in EFLAGS like interrupts IF (sti/cli), direction DF (std/cld), and alignment-check (stac/clac) have set/clear instructions, but not the condition flags ZF/SF/OF/PF or the BCD-carry AF.
cmp [rdi + 4 * rsi], edx will un-laminate even on Haswell/Skylake because of the indexed addressing mode, and it it doesn't have a read/write destination register (so it's not like add reg, [mem].)
If tuning only for Sandybridge-family, you might as well just increment a pointer and decrement the size counter. Although this does save back-end (unfused-domain) uops for RS-size effects.
In practice you'd want to unroll with a pointer increment.
You mentioned sizes from 0 to 32, so we need to skip the loop if RSI = 0. The code in your question is just a do{}while which doesn't do that. NEG sets flags according to the result, so we can JZ on that. You'd hope that it could macro-fuse because NEG is exactly like SUB from 0, but according to Agner Fog it doesn't on SnB/IvB. So that costs us another uop in the startup if you really do need to handle size=0.
Using integer registers
The standard way to implement integer += (a < b) or any other flag condition is what compilers do (Godbolt):
xor edx,edx ; can be hoisted out of a short-running loop, but compilers never do that
; but an interrupt-handler will destroy the rdx=dl status
cmp/test/whatever ; flag-setting code here
setcc dl ; zero-extended to a full register because of earlier xor-zeroing
add eax, edx
Sometimes compilers (especially gcc) will use setcc dl / movzx edx,dl, which puts the MOVZX on the critical path. This is bad for latency, and mov-elimination doesn't work on Intel CPUs when they use (part of) the same register for both operands.
For small arrays, if you don't mind having only an 8-bit counter, you could just use 8-bit add so you don't have to worry about zero-extension inside the loop.
; slower than cmp/adc: 5 uops per iteration so you'll definitely want to unroll.
; requires size<256 or the count will wrap
; use the add eax,edx version if you need to support larger size
count_signed_lt: ; (int *arr, size_t size, int key)
xor eax, eax
lea rdi, [rdi + rsi*4]
neg rsi ; we loop from -length to zero
jz .return ; if(-size == 0) return 0;
; xor edx, edx ; tmp destination for SETCC
.loop:
cmp [rdi + 4 * rsi], edx
setl dl ; false dependency on old RDX on CPUs other than P6-family
add al, dl
; add eax, edx ; boolean condition zero-extended into RDX if it was xor-zeroed
inc rsi
jnz .loop
.return:
ret
Alternatively using CMOV, making the loop-carried dep chain 2 cycles long (or 3 cycles on Intel before Broadwell, where CMOV is 2 uops):
;; 3 uops without any partial-register shenanigans, (or 4 because of unlamination)
;; but creates a 2 cycle loop-carried dep chain
cmp [rdi + 4 * rsi], edx
lea ecx, [rax + 1] ; tmp = count+1
cmovl eax, ecx ; count = arr[i]<key ? count+1 : count
So at best (with loop unrolling and a pointer-increment allowing cmp to micro-fuse) this takes 3 uops per element instead of 2.
SETCC is a single uop, so this is 5 fused-domain uops inside the loop. That's much worse on Sandybridge/IvyBridge, and still runs at worse than 1 per clock on later SnB-family. (Some ancient CPUs had slow setcc, like Pentium 4, but it's efficient on everything we still care about.)
When unrolling, if you want this to run faster than 1 cmp per clock, you have two choices: use separate registers for each setcc destination, creating multiple dep chains for the false dependencies, or use one xor edx,edx inside the loop to break the loop-carried false dependency into multiple short dep chains that only couple the setcc results of nearby loads (probably coming from the same cache line). You'll also need multiple accumulators because add latency is 1c.
Obviously you'll need to use a pointer-increment so cmp [rdi], edx can micro-fuse with a non-indexed addressing mode, otherwise the cmp/setcc/add is 4 uops total, and that's the pipeline width on Intel CPUs.
There's no partial-register stall from the caller reading EAX after writing AL, even on P6-family, because we xor-zeroed it first. Sandybridge won't rename it separately from RAX because add al,dl is a read-modify-write, and IvB and later never rename AL separately from RAX (only AH/BH/CH/DH). CPUs other than P6 / SnB-family don't do partial-register renaming at all, only partial flags.
The same applies for the version that reads EDX inside the loop. But an interrupt-handler saving/restoring RDX with push/pop would destroy its xor-zeroed status, leading to partial-register stalls every iteration on P6-family. This is catastrophically bad, so that's one reason compilers never hoist the xor-zeroing. They usually don't know if a loop will be long-running or not, and won't take the risk. By hand, you'd probably want to unroll and xor-zero once per unrolled loop body, rather than once per cmp/setcc.
You can use SSE2 or MMX for scalar stuff
Both are baseline on x86-64. Since you're not gaining anything (on SnB-family) from folding the load into the cmp, you might as well use a scalar movd load into an XMM register. MMX has the advantage of smaller code-size, but requires EMMS when you're done. It also allows unaligned memory operands, so it's potentially interesting for simpler auto-vectorization.
Until AVX512, we only have comparison for greater-than available, so it would take an extra movdqa xmm,xmm instruction to do key > arr[i] without destroying key, instead of arr[i] > key. (This is what gcc and clang do when auto-vectorizing).
AVX would be nice, for vpcmpgtd xmm0, xmm1, [rdi] to do key > arr[i], like gcc and clang use with AVX. But that's a 128-bit load, and we want to keep it simple and scalar.
We can decrement key and use (arr[i]<key) = (arr[i] <= key-1) = !(arr[i] > key-1). We can count elements where the array is greater-than key-1, and subtract that from the size. So we can make do with just SSE2 without costing extra instructions.
If key was already the most-negative number (so key-1 would wrap), then no array elements can be less than it. This does introduce a branch before the loop if that case is actually possible.
; signed version of the function in your question
; using the low element of XMM vectors
count_signed_lt: ; (int *arr, size_t size, int key)
; actually only works for size < 2^32
dec edx ; key-1
jo .key_eq_int_min
movd xmm2, edx ; not broadcast, we only use the low element
movd xmm1, esi ; counter = size, decrement toward zero on elements >= key
;; pxor xmm1, xmm1 ; counter
;; mov eax, esi ; save original size for a later SUB
lea rdi, [rdi + rsi*4]
neg rsi ; we loop from -length to zero
.loop:
movd xmm0, [rdi + 4 * rsi]
pcmpgtd xmm0, xmm2 ; xmm0 = arr[i] gt key-1 = arr[i] >= key = not less-than
paddd xmm1, xmm0 ; counter += 0 or -1
;; psubd xmm1, xmm0 ; -0 or -(-1) to count upward
inc rsi
jnz .loop
movd eax, xmm1 ; size - count(elements > key-1)
ret
.key_eq_int_min:
xor eax, eax ; no array elements are less than the most-negative number
ret
This should be the same speed as your loop on Intel SnB-family CPUs, plus a tiny bit of extra overhead outside. It's 4 fuse-domain uops, so it can issue at 1 per clock. A movd load uses a regular load port, and there are at least 2 vector ALU ports that can run PCMPGTD and PADDD.
Oh, but on IvB/SnB the macro-fused inc/jnz requires port 5, while PCMPGTD / PADDD both only run on p1/p5, so port 5 throughput will be a bottleneck. On HSW and later the branch runs on port 6, so we're fine for back-end throughput.
It's worse on AMD CPUs where a memory-operand cmp can use an indexed addressing mode without a penalty. (And on Intel Silvermont, and Core 2 / Nehalem, where memory-source cmp can be a single uop with an indexed addressing mode.)
And on Bulldozer-family, a pair of integer cores share a SIMD unit, so sticking to integer registers could be an even bigger advantage. That's also why int<->XMM movd/movq has higher latency, again hurting this version.
Other tricks:
Clang for PowerPC64 (included in the Godbolt link) shows us a neat trick: zero or sign-extend to 64-bit, subtract, and then grab the MSB of the result as a 0/1 integer that you add to counter. PowerPC has excellent bitfield instructions, including rldicl. In this case, it's being used to rotate left by 1, and then zero all bits above that, i.e. extracting the MSB to the bottom of another register. (Note that PowerPC documentation numbers bits with MSB=0, LSB=63 or 31.)
If you don't disable auto-vectorization, it uses Altivec with a vcmpgtsw / vsubuwm loop, which I assume does what you'd expect from the names.
# PowerPC64 clang 9-trunk -O3 -fno-tree-vectorize -fno-unroll-loops -mcpu=power9
# signed int version
# I've added "r" to register names, leaving immediates alone, because clang doesn't have `-mregnames`
... setup
.LBB0_2: # do {
lwzu r5, 4(r6) # zero-extending load and update the address register with the effective-address. i.e. pre-increment
extsw r5, r5 # sign-extend word (to doubleword)
sub r5, r5, r4 # 64-bit subtract
rldicl r5, r5, 1, 63 # rotate-left doubleword immediate then clear left
add r3, r3, r5 # retval += MSB of (int64_t)arr[i] - key
bdnz .LBB0_2 # } while(--loop_count);
I think clang could have avoided the extsw inside the loop if it had used an arithmetic (sign-extending) load. The only lwa that updates the address register (saving an increment) seems to be the indexed form lwaux RT, RA, RB, but if clang put 4 in another register it could use it. (There doesn't seem to be a lwau instruction.) Maybe lwaux is slow or maybe it's a missed optimization. I used -mcpu=power9 so even though that instruction is POWER-only, it should be available.
This trick could sort of help for x86, at least for a rolled-up loop.
It takes 4 uops this way per compare, not counting loop overhead. Despite x86's pretty bad bitfield extract capabilities, all we actually need is a logical right-shift to isolate the MSB.
count_signed_lt: ; (int *arr, size_t size, int key)
xor eax, eax
movsxd rdx, edx
lea rdi, [rdi + rsi*4]
neg rsi ; we loop from -length to zero
.loop:
movsxd rcx, dword [rdi + 4 * rsi] ; 1 uop, pure load
sub rcx, rdx ; (int64_t)arr[i] - key
shr rcx, 63 ; extract MSB
add eax, ecx ; count += MSB of (int64_t)arr[i] - key
inc rsi
jnz .loop
ret
This doesn't have any false dependencies, but neither does 4-uop xor-zero / cmp / setl / add. The only advantage here is that this is 4 uops even with an indexed addressing mode. Some AMD CPUs may run MOVSXD through an ALU as well as a load port, but Ryzen has the same latency as for it as for regular loads.
If you have fewer than 64 iterations, you could do something like this if only throughput matters, not latency. (But you can probably still do better with setl)
.loop
movsxd rcx, dword [rdi + 4 * rsi] ; 1 uop, pure load
sub rcx, rdx ; (int64_t)arr[i] - key
shld rax, rcx, 1 ; 3 cycle latency
inc rsi / jnz .loop
popcnt rax, rax ; turn the bitmap of compare results into an integer
But the 3-cycle latency of shld makes this a showstopper for most uses, even though it's only a single uop on SnB-family. The rax->rax dependency is loop-carried.
There's a trick to convert a signed comparison to an unsigned comparison and vice versa by toggling the top bit
bool signedLessThan(int a, int b)
{
return ((unsigned)a ^ INT_MIN) < b; // or a + 0x80000000U
}
It works because the ranges in 2's complement are still linear, just with a swapped signed and unsigned space. So the simplest way may be XORing before comparison
xor eax, eax
xor edx, 0x80000000 ; adjusting the search value
lea rdi, [rdi + rsi*4] ; pointer to end of array = base + length
neg rsi ; we loop from -length to zero
loop:
mov ecx, [rdi + 4 * rsi]
xor ecx, 0x80000000
cmp ecx, edx
adc rax, 0 ; only a single uop on Sandybridge-family even before BDW
inc rsi
jnz loop
If you can modify the array then just do the conversion before checking
In ADX there's ADOX that uses carry from OF. Unfortunately signed comparison also needs SF instead of only OF, thus you can't use it like this
xor ecx, ecx
loop:
cmp [rdi + 4 * rsi], edx
adox rax, rcx ; rcx=0; ADOX is not available with an immediate operand
and must do some more bit manipulations to correct the result
In the case the array is guaranteed to be sorted, one could use cmovl with an "immediate" value representing the correct value to add. There are no immediates for cmovl, so you'll have to load them into registers beforehand.
This technique makes sense when unrolled, for example:
; load constants
mov r11, 1
mov r12, 2
mov r13, 3
mov r14, 4
loop:
xor ecx, ecx
cmp [rdi + 0], edx
cmovl rcx, r11
cmp [rdi + 4], edx
cmovl rcx, r12
cmp [rdi + 8], edx
cmovl rcx, r13
cmp [rdi + 12], edx
cmovl rcx, r14
add rax, rcx
; update rdi, test loop condition, etc
jcc loop
You have 2 uops per comparison, plus overhead. There is a 4-cycle (BDW and later) dependency chain between the cmovl instructions, but it is not carried.
One disadvantage is that you have to set up the 1,2,3,4 constants outside of the loop. It also doesn't work as well if not unrolled (you need to ammortize the add rax, rcx accumulation).
Assuming the array is sorted, you could make separate code branches for positive and negative needles. You will need a branch instruction at the very start, but after that, you can use the same branch-free implementation you would use for unsigned numbers. I hope that is acceptable.
needle >= 0:
go through the array in ascending order
begin by counting every negative array element
proceed with the positive numbers just like you would in the unsigned scenario
needle < 0:
go through the array in descending order
begin by skipping every positive array element
proceed with the negative numbers just like you would in the unsigned scenario
Unfortunately, in this approach you cannot unroll your loops.
An alternative would be to go through each array twice; once with the needle, and again to find the number of positive or negative elements (using a 'needle' that matches the minimum signed integer).
(unsigned) count the elements < needle
(unsigned) count the elements >= 0x80000000
add the results
if needle < 0, subtract the array length from the result
There's probably a lot to be optimized to my code below. I'm quite rusty at this.
; NOTE: no need to initialize eax here!
lea rdi, [rdi + rsi*4] ; pointer to end of array = base + length
neg rsi ; we loop from -length to zero
mov ebx, 80000000h ; minimum signed integer (need this in the loop too)
cmp edx, ebx ; set carry if needle negative
sbb eax, eax ; -1 if needle negative, otherwise zero
and eax, esi ; -length if needle negative, otherwise zero
loop:
cmp [rdi + 4 * rsi], edx
adc rax, 0 ; +1 if element < needle
cmp [rdi + 4 * rsi], ebx
cmc
adc rax, 0 ; +1 if element >= 0x80000000
inc rsi
jnz loop

what is fs:[register+value] meaning in assembly?

xor ebx,ebx
mov eax,DWORD PTR fs:[ebx+0x3]
I know first line of this code, but what's fs:[ebx+0x3]?
and why it giving me an error while compiling ?
test.asm:2: error: comma, colon, decorator or end of line expected after operand
The xor opcode sets EBX to 0. So the mov opcode accesses a DWORD at fs:[3]. This accesses the last byte of the Win32 Thread Information Block's Current Structured Exception Handling (SEH) frame located from fs:[0] to fs:[3] (4 bytes) and the first three bytes of the Stack Base variable, the initial value of ESP.
Unless you are confronted with some sophisticatedly constructed and obfuscated virus or anti-debugging-technique which combines these 3+1 bytes to something useful, this would just be a somewhat random number.
Another possibility is, that the "segment" register FS has been modified prior to these instructions to contain a sensible base address. In that case, this may be an useful instruction like many others. It is undecidable by the code snippet you provided.
For example
mov eax, fs
inc eax
mov fs, eax
...
xor ebx,ebx
mov eax,DWORD PTR fs:[ebx+0x3]
would return the 'Stack Base' in EAX - '(FS+1)+(0+3)' = real-FS:[4] = 'Stack Base' location.
Addition: For completeness: why it doesn't assemble has been mentioned in the comments by 'Ross Ridge': It's MASM syntax and not NASM syntax.

Assembly 8086: get and print 32bit filesize using 16bit registers

i need to print a filesize using function 23h http://www.ousob.com/ng/asm/ng4d85d.php but i dont understand how can i get value from (offset 21h) of the FCB, it should be 4byte value, so need to move it into pair of 16bit registers, e.g. BX and CX
this is my simple code:
mov dx,offset input
mov ah,23h
int 21h
then i need to print it, but i just know how to print one 16bit register and i dont know find some tutorial how to do this for 32bit value (2 registers) simply,
thanks
The FCB returns both the number of records (a 32-bit quantity at offset 21h of the FCB) and the size of each record (a 16-bit quantity at offset 0Eh of the FCB). You need to multiply those together to get an approximation of the file size. It's only an approximation because a partial block is counted as a whole block for this call.
So assuming that input is already declared as an FCB structure in memory in the Data Segment and that your file consists exclusively of records that 27 bytes long, your code might look like this.
mov dx,offset input ; point to FCB
mov si, dx ; copy pointer
add si,0Eh ; point to record size within FCB
mov [si],27 ; load record size = 27 bytes
mov ah,23h ; get file size
int 21h ; call the DOS interrupt
cmp al,0 ; was the call successful?
jnz error ; if not, handle the error
add si,13h ; advance pointer to point to FCB:21h
; now ds:si points to file size in records
See this derivation of the age-old "Ralf Brown's interrupt list" for a list of some of the many precautions regarding this interrupt.

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