汇编手写函数比GCC编译的函数慢。

huangapple go评论59阅读模式
英文:

Assembly handwritten function slower than GCC compiled function

问题

I decide to create a string-length function in Assembly (using FASM).
我决定在汇编中创建一个字符串长度函数(使用FASM)。

My function takes a string (no matter aligned at 8 bytes or not) and checks if it's aligned at 8 bytes. If it's aligned, the main process (loop) will be begun. Otherwise, first 8 characters will be checked one-by-one, then the string will be aligned at 8 bytes and continue ...
我的函数接受一个字符串(无论是否按8字节对齐),并检查它是否按8字节对齐。如果对齐了,主要过程(循环)将开始。否则,首先会逐个检查前8个字符,然后将字符串对齐到8字节并继续...

There will be no "end of the memory page" problem since the string will be aligned at 8 bytes boundary anyway and by this alignment, it will never face the end of memory page problem.
由于字符串无论如何都将对齐到8字节边界,所以不会出现“内存页面结束”的问题,通过这种对齐方式,它永远不会面临内存页面结束的问题。

But the problem is that I decided to implement its C version too, and I compiled it, and now I have 2 assembly codes, the one I wrote it and the one is written in C and compiled to assembly. The problem is the C version is up to 1.5x faster than my handwritten assembly !!!!!!! In my code, everything is just fine, and I even aligned the jump-points to 16 bytes and there is no nop running (except one, out of the loop which is kinda nothing (.align8 to .loop)) !!!
但问题是,我决定也要实现它的C版本,然后编译了它,现在我有两个汇编代码,一个是我自己写的,另一个是用C编写的并编译成汇编。问题是C版本比我手写的汇编快1.5倍!!!!在我的代码中,一切都很好,我甚至将跳转点对齐到16字节,并且没有nop运行(除了一个,在循环之外,这几乎什么都没有(从.align8.loop))!!!

I can't find why my pure assembly code is 1.5x slower than the GCC version !!!
我找不到为什么我的纯汇编代码比GCC版本慢1.5倍!!!

My Assembly source-code :
我的汇编源代码:

The GCC version :
GCC版本:

My function test result :
我的函数测试结果:

string length => 336
字符串长度 => 336
loop execution times => 10000000
循环执行次数 => 10000000
total execution time => 0.772015
总执行时间 => 0.772015

GCC function test result :
GCC函数测试结果:

string length => 336
字符串长度 => 336
loop execution times => 10000000
循环执行次数 => 10000000
total execution time => 0.522015
总执行时间 => 0.522015

What is the problem ? Why my function is 1.5x slower when everything is kinda looks fine?
问题是什么?为什么我的函数要慢1.5倍,当一切似乎都很正常?

My string is aligned at 8 bytes, so you can skip the first one-by-one process and alignment.
我的字符串对齐到了8字节,所以你可以跳过第一个逐个处理和对齐的过程。

Is there any problem with my label aligning ? Or the problem is from somewhere else?
我的标签对齐有问题吗?还是问题来自其他地方?

ABI -> x64 (Windows)
ABI -> x64(Windows)

CPU (Test) -> i7-7800X
CPU(测试)-> i7-7800X

My C test application source-code :
我的C测试应用程序源代码:

My object file (with these 2 slen functions to link to that C tester) creator in FASM :
我的目标文件(带有这两个slen函数,用于链接到C测试程序)是在FASM中创建的:

Also the C version of slen
还有slen的C版本

英文:

I decide to create a string-length function in Assembly (using FASM).
My function takes a string (no matter aligned at 8 bytes or not) and checks if it's aligned at 8 bytes. If it's aligned, the main process (loop) will be begun. Otherwise, first 8 characters will be checked one-by-one, then the string will be aligned at 8 bytes and continue ...
There will be no "end of the memory page" problem since the string will be aligned at 8 bytes boundary anyway and by this alignment, it will never face the end of memory page problem.

But the problem is that I decided to implement its C version too, and I compiled it, and now I have 2 assembly codes, the one I wrote it and the one is written in C and compiled to assembly. The problem is the C version is up to 1.5x faster than my handwritten assembly !!!!!!! In my code, everything is just fine, and I even aligned the jump-points to 16 bytes and there is no nop running (except one, out of the loop which is kinda nothing (.align8 to .loop)) !!!
I can't find why my pure assembly code is 1.5x slower than the GCC version !!!

My Assembly source-code :

 align 16
slen:
        mov     r8, rcx
        test    cl, 7
        jz      .loop
        xor     eax, eax
        cmp     BYTE [rcx], al
        je      SHORT .ret
        cmp     BYTE [rcx+1], al
        je      SHORT .ret1
        cmp     BYTE [rcx+2], al
        je      SHORT .ret2
        cmp     BYTE [rcx+3], al
        je      SHORT .ret3
        cmp     BYTE [rcx+4], al
        je      SHORT .ret4
        cmp     BYTE [rcx+5], al
        je      SHORT .ret5
        cmp     BYTE [rcx+6], al
        je      SHORT .ret6
        cmp     BYTE [rcx+7], al
        jne     SHORT .align8
        mov     al, 7
        ret
 align 16
 .ret:  ret
 align 16
 .ret1: mov     al, 1
        ret
 align 16
 .ret2: mov     al, 2
        ret
 align 16
 .ret3: mov     al, 3
        ret
 align 16
 .ret4: mov     al, 4
        ret
 align 16
 .ret5: mov     al, 5
        ret
 align 16
 .ret6: mov     al, 6
        ret
 align 16
 .align8:
        lea     rcx, [rcx+7]
        and     rcx, (-8)
 align 16
 .loop: mov     rax, QWORD [rcx]
        test    al, al
        jz      SHORT .end
        test    ah, ah
        jz      SHORT .end.1
        test    eax, 0x00ff0000
        jz      SHORT .end.2
        test    eax, 0xff000000
        jz      SHORT .end.3
        shr     rax, 32
        test    al, al
        jz      SHORT .end.4
        test    ah, ah
        jz      SHORT .end.5
        test    eax, 0x00ff0000
        jz      SHORT .end.6
        test    eax, 0xff000000
        jz      SHORT .end.7
        add     rcx, 8
        jmp     SHORT .loop
 align 16
 .end: mov      rax, rcx
        sub     rax, r8
        ret
 align 16
 .end.1:
        lea     rax, [rcx+1]
        sub     rax, r8
        ret
 .end.2:
        lea     rax, [rcx+2]
        sub     rax, r8
        ret
 .end.3:
        lea     rax, [rcx+3]
        sub     rax, r8
        ret
 .end.4:
        lea     rax, [rcx+4]
        sub     rax, r8
        ret
 .end.5:
        lea     rax, [rcx+5]
        sub     rax, r8
        ret
 .end.6:
        lea     rax, [rcx+6]
        sub     rax, r8
        ret
 .end.7:
        lea     rax, [rcx+7]
        sub     rax, r8
        ret       

The GCC version :

 align 16
slen:
        test    cl, 7
        je      .L18
        xor     eax, eax
        cmp     BYTE [rcx], 0
        je      .L1
        cmp     BYTE [rcx+1], 0
        mov     eax, 1
        je      .L1
        cmp     BYTE [rcx+2], 0
        mov     eax, 2
        je      .L1
        cmp     BYTE [rcx+3], 0
        mov     eax, 3
        je      .L1
        cmp     BYTE [rcx+4], 0
        mov     eax, 4
        je      .L1
        cmp     BYTE [rcx+5], 0
        mov     eax, 5
        je      .L1
        cmp     BYTE [rcx+6], 0
        mov     eax, 6
        je      .L1
        cmp     BYTE [rcx+7], 0
        mov     eax, 7
        je      .L1
        lea     rax, [rcx+7]
        and     rax, -8
        jmp     .L47
 align 16
.L18:
        mov     rax, rcx
        jmp     .L47
 align 16
.L40:
        test    dh, dh
        je      .L49
        test    edx, 16711680
        je      .L50
        test    edx, 4278190080
        je      .L51
        shr     rdx, 32
        test    dl, dl
        je      .L52
        test    dh, dh
        je      .L53
        test    edx, 16711680
        je      .L54
        test    edx, 4278190080
        je      .L55
        add     rax, 8
.L47:
        mov     rdx, QWORD [rax]
        test    dl, dl
        jne     .L40
        sub     eax, ecx
.L1:
        ret
 align 16
.L49:
        sub     rax, rcx
        add     eax, 1
        ret
 align 16
.L50:
        sub     rax, rcx
        add     eax, 2
        ret
 align 16
.L51:
        sub     rax, rcx
        add     eax, 3
        ret
 align 16
.L52:
        sub     rax, rcx
        add     eax, 4
        ret
 align 16
.L53:
        sub     rax, rcx
        add     eax, 5
        ret
 align 16
.L54:
        sub     rax, rcx
        add     eax, 6
        ret
 align 16
.L55:
        sub     rax, rcx
        add     eax, 7
        ret   

My function test result :

string length => 336
loop execution times => 10000000
total execution time => 0.772015

GCC function test result :

string length => 336
loop execution times => 10000000
total execution time => 0.522015

What is the problem ? Why my function is 1.5x slower when everything is kinda looks fine?
My string is aligned at 8 bytes, so you can skip the first one-by-one process and alignment.

Is there any problem with my label aligning ? Or the problem is from somewhere else?

ABI -> x64 (Windows)

CPU (Test) => i7-7800X

My C test application source-code :

#include <stdio.h>
#include <stdlib.h>
#include <windows.h>

unsigned int
slen_by_me(const char *);

unsigned int
slen_gcc(const char *);

int main() {
    static const char *str="WWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWW";
    LARGE_INTEGER frequency;
    LARGE_INTEGER start;
    LARGE_INTEGER end;
    double interval;
    unsigned int l = 0;

    QueryPerformanceFrequency(&frequency);
    QueryPerformanceCounter(&start);

    for (int i = 0; i < 10000000; i++) {
        l += slen_gcc(str);
    }

    QueryPerformanceCounter(&end);
    interval = (double) (end.QuadPart - start.QuadPart) / frequency.QuadPart;

    printf("%f\n%u\n", interval, l);
    return 0;
}

My object file (with these 2 slen functions to link to that C tester) creator in FASM :

format MS64 COFF

public slen_gcc
public slen_by_me

section '.text' code readable executable align 64

 align 16
slen_gcc:
        test    cl, 7
        je      .L18
        xor     eax, eax
        cmp     BYTE [rcx], 0
        je      .L1
        cmp     BYTE [rcx+1], 0
        mov     eax, 1
        je      .L1
        cmp     BYTE [rcx+2], 0
        mov     eax, 2
        je      .L1
        cmp     BYTE [rcx+3], 0
        mov     eax, 3
        je      .L1
        cmp     BYTE [rcx+4], 0
        mov     eax, 4
        je      .L1
        cmp     BYTE [rcx+5], 0
        mov     eax, 5
        je      .L1
        cmp     BYTE [rcx+6], 0
        mov     eax, 6
        je      .L1
        cmp     BYTE [rcx+7], 0
        mov     eax, 7
        je      .L1
        lea     rax, [rcx+7]
        and     rax, -8
        jmp     .L47
 align 16
.L18:
        mov     rax, rcx
        jmp     .L47
 align 16
.L40:
        test    dh, dh
        je      .L49
        test    edx, 16711680
        je      .L50
        test    edx, 4278190080
        je      .L51
        shr     rdx, 32
        test    dl, dl
        je      .L52
        test    dh, dh
        je      .L53
        test    edx, 16711680
        je      .L54
        test    edx, 4278190080
        je      .L55
        add     rax, 8
.L47:
        mov     rdx, QWORD [rax]
        test    dl, dl
        jne     .L40
        sub     eax, ecx
.L1:
        ret
 align 16
.L49:
        sub     rax, rcx
        add     eax, 1
        ret
 align 16
.L50:
        sub     rax, rcx
        add     eax, 2
        ret
 align 16
.L51:
        sub     rax, rcx
        add     eax, 3
        ret
 align 16
.L52:
        sub     rax, rcx
        add     eax, 4
        ret
 align 16
.L53:
        sub     rax, rcx
        add     eax, 5
        ret
 align 16
.L54:
        sub     rax, rcx
        add     eax, 6
        ret
 align 16
.L55:
        sub     rax, rcx
        add     eax, 7
        ret

 align 16
slen_by_me:
        mov     r8, rcx
        test    cl, 7
        jz      .loop
        xor     eax, eax
        cmp     BYTE [rcx], al
        je      SHORT .ret
        cmp     BYTE [rcx+1], al
        je      SHORT .ret1
        cmp     BYTE [rcx+2], al
        je      SHORT .ret2
        cmp     BYTE [rcx+3], al
        je      SHORT .ret3
        cmp     BYTE [rcx+4], al
        je      SHORT .ret4
        cmp     BYTE [rcx+5], al
        je      SHORT .ret5
        cmp     BYTE [rcx+6], al
        je      SHORT .ret6
        cmp     BYTE [rcx+7], al
        jne     SHORT .align8
        mov     al, 7
        ret
 align 16
 .ret:  ret
 align 16
 .ret1: mov     al, 1
        ret
 align 16
 .ret2: mov     al, 2
        ret
 align 16
 .ret3: mov     al, 3
        ret
 align 16
 .ret4: mov     al, 4
        ret
 align 16
 .ret5: mov     al, 5
        ret
 align 16
 .ret6: mov     al, 6
        ret
 align 16
 .align8:
        lea     rcx, [rcx+7]
        and     rcx, (-8)
 align 16
 .loop: mov     rax, QWORD [rcx]
        test    al, al
        jz      SHORT .end
        test    ah, ah
        jz      SHORT .end.1
        test    eax, 0x00ff0000
        jz      SHORT .end.2
        test    eax, 0xff000000
        jz      SHORT .end.3
        shr     rax, 32
        test    al, al
        jz      SHORT .end.4
        test    ah, ah
        jz      SHORT .end.5
        test    eax, 0x00ff0000
        jz      SHORT .end.6
        test    eax, 0xff000000
        jz      SHORT .end.7
        add     rcx, 8
        jmp     SHORT .loop
 align 16
 .end: mov      rax, rcx
        sub     rax, r8
        ret
 align 16
 .end.1:
        lea     rax, [rcx+1]
        sub     rax, r8
        ret
 .end.2:
        lea     rax, [rcx+2]
        sub     rax, r8
        ret
 .end.3:
        lea     rax, [rcx+3]
        sub     rax, r8
        ret
 .end.4:
        lea     rax, [rcx+4]
        sub     rax, r8
        ret
 .end.5:
        lea     rax, [rcx+5]
        sub     rax, r8
        ret
 .end.6:
        lea     rax, [rcx+6]
        sub     rax, r8
        ret
 .end.7:
        lea     rax, [rcx+7]
        sub     rax, r8
        ret

Also the C version of slen

int
slen(const char *str) {
    const char *start=str;
    if(((unsigned long long)str & 7) != 0) {
        if(str[0] == 0x00)
            return 0;
        if(str[1] == 0x00)
            return 1;
        if(str[2] == 0x00)
            return 2;
        if(str[3] == 0x00)
            return 3;
        if(str[4] == 0x00)
            return 4;
        if(str[5] == 0x00)
            return 5;
        if(str[6] == 0x00)
            return 6;
        if(str[7] == 0x00)
            return 7;
        str=(const char *)(((unsigned long long)str + 7) & (-8));
    }
    do {
        unsigned long long bytes=(*(unsigned long long*)(str));
        if((unsigned char)bytes==0x00)
            return (int)(str-start);
        if((bytes & 0x0000ff00)==0)
            return (int)(str-start+1);
        if((bytes & 0x00ff0000)==0)
            return (int)(str-start+2);
        if((bytes & 0xff000000)==0)
            return (int)(str-start+3);
        bytes >>= 32;
        if((unsigned char)bytes==0x00)
            return (int)(str-start+4);
        if((bytes & 0x0000ff00)==0)
            return (int)(str-start+5);
        if((bytes & 0x00ff0000)==0)
            return (int)(str-start+6);
        if((bytes & 0xff000000)==0)
            return (int)(str-start+7);
        str+=8;
    } while (1);
}

答案1

得分: 3

请问您需要哪部分内容进行翻译?

英文:

Allow me to refer you to one of my Pure Assembly library function (coming soon).
According to your question, it's about strlen (which named "str_length" in my library and developed for both Microsoft x64 ABI and System-v AMD64 ABI).

I remember (a few years ago) there was a C/C++ function about this type of string length calculator function.

size_t my_strlen(const char *s) {
    size_t len = 0;
    for(;;) {
        unsigned x = *(unsigned*)s;
        if((x & 0xFF) == 0) return len;
        if((x & 0xFF00) == 0) return len + 1;
        if((x & 0xFF0000) == 0) return len + 2;
        if((x & 0xFF000000) == 0) return len + 3;
        s += 4, len += 4;
    }
}

Even named "FAST strlen" which it's really not that fast. So, i decided to write my own "FAST strlen" in Assembly.

In x86-64, it's possible to load a 8-BYTE chunk into a 64-bit register so why 4-BYTE loading ? (As 'size_t my_strlen(const char *s)' did)

JCC Erratum

About 'JCC Erratum', still there are too many Skylake CPUs in the world (and by world, i mean DataCenters (check Hetzner datacenter and you find too many Skylake and old CPUs)). It's not optional, you MUST take care of this bad boy. But, it's very important to handle it without adding a NOP or even prefixes. Because by doing this, you make new problems for other CPUs. You can handle it by creating small new branches and putting some codes into a fresh 32-BYTE chunk (But don't make it too heavy).

TAKE CARE OF LOOP TAIL JUMP

Another subject, is taking care about the loops tail jump. Also again, you MUST make a tail for your loop and using jmp (unconditional jump) to jump to that tail (because of predictable branches subject (read the Agner Fog document about this bad boy (I love this guy for no reason xD)). Also, As Mr. Peter Cordes mentioned, and if you check the GCC jump method, you find the solution about loop creation and jumps.

TAKE CARE OF BRANCH ALIGNMENT

Yes, take care of branch alignment (16-BYTE boundaries), specially those you jump to, too many times (well, good boy|girl (sorry, no name), it's handled by you).

GCC REALLY ?! WHY NOT MACRO-FUSED ?

Well you (the question starter) did a right thing. You used a register for unaligned condition cmp so you have the benefit of macro-fusing. But in the code generated by GCC, you can see that cmp BYTE PTR [rcx], 0 is used. This will removes the benefit of macro-fusing from your code (its code actually (GCC)).
Of course, GCC done it to handle the padding but it's really not acceptable.

An example of this situation in uiCA test tool:

0000000000000000 <.text>:
   0:	80 39 00             	cmp    BYTE PTR [rcx],0x0
   3:	0f 84 00 00 00 00    	je     0x9
   9:	38 01                	cmp    BYTE PTR [rcx],al
   b:	0f 84 00 00 00 00    	je     0x11

The second cmp got M flag which stands for 'Macro-fused with previous instruction'.

> Macro Fusion is restricted to 16-bit and 32-bit mode only (including
> 32-bit compatibility sub-mode in x86-64). CMP and TEST can fuse when
> comparing:
>
> REG-REG. (e.g, CMP EAX,ECX; JZ label)
> REG-IMM. (e.g., CMP EAX,0x80; JZ label)
> REG-MEM. (e.g., CMP EAX,[ECX]; JZ label)
> MEM-REG. (e.g., CMP [EAX],ECX; JZ label)
>
> CMP and TEST can not be fused when comparing MEM-IMM (e.g. CMP
> [EAX],0x80; JZ label)

And finally about the function and its performance test (according to your needs). I bring you the one with Microsoft x64 ABI.

; libASM, independent standard libraries in Assembly (programming-language).
; For more information, please visit the libASM website (www.libasm.com).
; Copyright (C) 2023 Mr. Alireza Saeidipour. All rights reserved.

; Published by SOURCEBRING, under its international legal terms and conditions.
; For more information, please visit the SOURCEBRING website (www.sourcebring.com).

; “FAILURE GUARANTEES SUCCESS”
; — Alireza Saeidipour

	align.function
str_length:
	mov	r8, rcx
	test	cl, 7
	jz	@f
	xor	eax, eax
	cmp	BYTE [rcx], al
	je	SHORT .len0
	cmp	BYTE [rcx+1], al
	jne	SHORT .unaligned_continue
	mov	al, 1
	ret
	align.branch32
 .unaligned_continue:
	cmp	BYTE [rcx+2], al
	je	SHORT .len2
	cmp	BYTE [rcx+3], al
	je	SHORT .len3
	cmp	BYTE [rcx+4], al
	je	SHORT .len4
	cmp	BYTE [rcx+5], al
	je	SHORT .len5
	cmp	BYTE [rcx+6], al
	je	.len6
	cmp	BYTE [rcx+7], al
	je	.len7
	lea	r8, [rcx+7]
	and	r8, (-8)
	jmp	@f
	align.branch
 .len0:	ret
	align.branch
 .len2:	mov	eax, 2
	ret
	align.branch
 .len3:	mov	eax, 3
	ret
	align.branch
 .len4:	mov	eax, 4
	ret
	align.branch
 .len5:	mov	eax, 5
	ret
	align.branch
 .len6:	mov	eax, 6
	ret
	align.branch
 .len7:	mov	eax, 7
	ret
	align.branch
 .return_add7:
	lea	rax, [r8+7]
	sub	rax, r9
	ret
	align.branch
 @@:	mov	r9, rcx
	mov	ecx, 0x00ff0000
	mov	edx, 0xff000000
	jmp	SHORT @f
	align.branch32
 .loop:	test	eax, ecx
	jz	SHORT .return_add2
	test	eax, edx
	jz	SHORT .return_add3
	shr	rax, 32
	test	al, al
	jz	SHORT .return_add4
	test	ah, ah
	jz	SHORT .return_add5
	test	eax, ecx
	jz	SHORT .return_add6
	test	eax, edx
	jz	SHORT .return_add7
	add	r8, 8
 @@:	mov	rax, QWORD [r8]
	test	al, al
	jz	SHORT .return
	test	ah, ah
	jnz	SHORT .loop
	lea	rax, [r8+1]
	sub	rax, r9
	ret
	align.branch
 .return:
	mov	rax, r8
	sub	rax, r9
	ret
	align.branch
 .return_add2:
	lea	rax, [r8+2]
	sub	rax, r9
	ret
	align.branch
 .return_add3:
	lea	rax, [r8+3]
	sub	rax, r9
	ret
	align.branch
 .return_add4:
	lea	rax, [r8+4]
	sub	rax, r9
	ret
	align.branch
 .return_add5:
	lea	rax, [r8+5]
	sub	rax, r9
	ret
	align.branch
 .return_add6:
	lea	rax, [r8+6]
	sub	rax, r9
	ret
 .size = $ - str_length

And Macros in this source-code:

macro align.function { align 32 }
macro align.branch { align 16 }
macro align.branch32 { align 32 }

This function considered as high-end solution (non-SIMD). You can find SIMD version of this function (9 functions are created only for string length operation) in my library soon (The library will be released by the end of June (2023)).

str_length
str_length_sse2
str_length_avx
str_length_avx2
str_length_avx512bw
str_length_long_sse2
str_length_long_avx
str_length_long_avx2
str_length_long_avx512bw

Test results (based on your parameters and your test tools (function (C)):

string length => 336
loop execution times => 10000000
total execution time => 0.430173

Yes, even faster than the one generated by GCC (0.522015). You will get same result for an unaligned string too.

Also, there is no 'JCC Erratum' problem in my code (The hex string of my function for you to check it).

 49 89 c8 f6 c1 07 0f 84 c4 00 00 00 31 c0 38 01
 74 3e 38 41 01 75 09 b0 01 c3 90 90 90 90 90 90
 38 41 02 74 3b 38 41 03 74 46 38 41 04 74 51 38
 41 05 74 5c 38 41 06 74 67 38 41 07 74 72 4c 8d
 41 07 49 83 e0 f8 e9 85 00 00 00 90 90 90 90 90
 c3 90 90 90 90 90 90 90 90 90 90 90 90 90 90 90
 b8 02 00 00 00 c3 90 90 90 90 90 90 90 90 90 90
 b8 03 00 00 00 c3 90 90 90 90 90 90 90 90 90 90
 b8 04 00 00 00 c3 90 90 90 90 90 90 90 90 90 90
 b8 05 00 00 00 c3 90 90 90 90 90 90 90 90 90 90
 b8 06 00 00 00 c3 90 90 90 90 90 90 90 90 90 90
 b8 07 00 00 00 c3 90 90 90 90 90 90 90 90 90 90
 49 8d 40 07 4c 29 c8 c3 90 90 90 90 90 90 90 90
 49 89 c9 b9 00 00 ff 00 ba 00 00 00 ff eb 21 90
 85 c8 74 4c 85 d0 74 58 48 c1 e8 20 84 c0 74 60
 84 e4 74 6c 85 c8 74 78 85 d0 74 c4 49 83 c0 08
 49 8b 00 84 c0 74 19 84 e4 75 d5 49 8d 40 01 4c
 29 c8 c3 90 90 90 90 90 90 90 90 90 90 90 90 90
 4c 89 c0 4c 29 c8 c3 90 90 90 90 90 90 90 90 90
 49 8d 40 02 4c 29 c8 c3 90 90 90 90 90 90 90 90
 49 8d 40 03 4c 29 c8 c3 90 90 90 90 90 90 90 90
 49 8d 40 04 4c 29 c8 c3 90 90 90 90 90 90 90 90
 49 8d 40 05 4c 29 c8 c3 90 90 90 90 90 90 90 90
 49 8d 40 06 4c 29 c8 c3

Warning: Please attention that FASM uses too many NOP for 'align' directive (instead of using a long NOP) so don't use this directive when there is no jmp above of it (As you say, direct access).

Warning: For old CPUs sake, keep your jumps body short and use registers instead of imm. And always handle 'JCC Erratum' (You lose 1.3x performance for that).

With best-regards.

答案2

得分: 2

I changed my code from:

.loop: mov     rax, QWORD [rcx]
        test    al, al
        jz      SHORT .end
        test    ah, ah
        jz      SHORT .end.1
        test    eax, 0x00ff0000
        jz      SHORT .end.2
        test    eax, 0xff000000
        jz      SHORT .end.3
        shr     rax, 32
        test    al, al
        jz      SHORT .end.4
        test    ah, ah
        jz      SHORT .end.5
        test    eax, 0x00ff0000
        jz      SHORT .end.6
        test    eax, 0xff000000
        jz      SHORT .end.7
        add     rcx, 8
        jmp     SHORT .loop 

To (first, we jump to the '.loop' label):

.loop.continue:
        test    ah, ah
        jz      SHORT .end1
        test    eax, 0x00ff0000
        jz      SHORT .end2
        test    eax, 0xff000000
        jz      SHORT .end3
        shr     rax, 32
        test    al, al
        jz      SHORT .end4
        test    ah, ah
        jz      SHORT .end5
        test    eax, 0x00ff0000
        jz      SHORT .end6
        test    eax, 0xff000000
        jz      .end7
        lea     rcx, [rcx+8]
 .loop: mov     rax, QWORD [rcx]
        test    al, al
        jnz     SHORT .loop.continue
        mov     rax, rcx
        sub     rax, rdx
        ret

And even with 'JCC Erratum' problem, I get amazing result (0.532015).
There was something wrong with my loop. In the first one, we jumped to the loop, a QWORD was taken, and we started to search for 0x00, and at the end of the loop, 8 was added to rcx (string memory address), and we had to jump to the loop (top) again.

But in the solution, we jump to the end of the loop and handle the first check, then we jump back to handle the others, and by doing this, the speed problem is fixed !!!

UPDATED

I just tried to make the loop body smaller (in size, like my first code), and the result was amazing:

strl:
        push    rdi
        push    rsi
        mov     rdi, rcx
        mov     rsi, rcx
        mov     ecx, 0x00ff0000
        mov     edx, 0xff000000
        mov     r8, 0x000000ff00000000
        mov     r9, 0x0000ff0000000000
        mov     r10, 0x00ff000000000000
        mov     r11, 0xff00000000000000
        test    dil, 7
        jz      @f
        ; handle unaligned
  align 32
 @@:    mov     rax, QWORD [rdi]
        test    al, al
        jz      SHORT .end
        test    ah, ah
        jz      SHORT .end1
        test    eax, ecx
        jz      SHORT .end2
        test    eax, edx
        jz      SHORT .end3
        test    rax, r8
        jz      SHORT .end4
        test    rax, r9
        jz      SHORT .end5
        test    rax, r10
        jz      SHORT .end6
        test    rax, r11
        jz      SHORT .end7
        add     rdi, 8
        jmp     @b
  align 16
 .end:  mov     rax, rdi
        sub     rax, rsi
        pop     rsi
        pop     rdi
        ret
  align 16
 .end1: lea     rax, [rdi+1]
        sub     rax, rsi
        pop     rsi
        pop     rdi
        ret
  align 16
 .end2: lea     rax, [rdi+2]
        sub     rax, rsi
        pop     rsi
        pop     rdi
        ret
  align 16
 .end3: lea     rax, [rdi+3]
        sub     rax, rsi
        pop     rsi
        pop     rdi
        ret
  align 16
 .end4: lea     rax, [rdi+4]
        sub     rax, rsi
        pop     rsi
        pop     rdi
        ret
  align 16
 .end5: lea     rax, [rdi+5]
        sub     rax, rsi
        pop     rsi
        pop     rdi
        ret
  align 16
 .end6: lea     rax, [rdi+6]
        sub     rax, rsi
        pop     rsi
        pop     rdi
        ret
  align 16
 .end7: lea     rax, [rdi+7]
        sub     rax, rsi
        pop     rsi
        pop     rdi
        ret
英文:

I changed my code from

.loop: mov     rax, QWORD [rcx]
        test    al, al
        jz      SHORT .end
        test    ah, ah
        jz      SHORT .end.1
        test    eax, 0x00ff0000
        jz      SHORT .end.2
        test    eax, 0xff000000
        jz      SHORT .end.3
        shr     rax, 32
        test    al, al
        jz      SHORT .end.4
        test    ah, ah
        jz      SHORT .end.5
        test    eax, 0x00ff0000
        jz      SHORT .end.6
        test    eax, 0xff000000
        jz      SHORT .end.7
        add     rcx, 8
        jmp     SHORT .loop 

To (first, we jump to the '.loop' label):

.loop.continue:
        test    ah, ah
        jz      SHORT .end1
        test    eax, 0x00ff0000
        jz      SHORT .end2
        test    eax, 0xff000000
        jz      SHORT .end3
        shr     rax, 32
        test    al, al
        jz      SHORT .end4
        test    ah, ah
        jz      SHORT .end5
        test    eax, 0x00ff0000
        jz      SHORT .end6
        test    eax, 0xff000000
        jz      .end7
        lea     rcx, [rcx+8]
 .loop: mov     rax, QWORD [rcx]
        test    al, al
        jnz     SHORT .loop.continue
        mov     rax, rcx
        sub     rax, rdx
        ret

And even with 'JCC Erratum' problem, I get amazing result (0.532015).
There was something wrong with my loop. In the first one, we jumped to loop and a QWORD taken and we started to search for 0x00 and at the end of the loop, 8 added to rcx (string memory address) and we have to jump to the loop (top) again.

But in solution, we jump to the end of loop and we handle the first check then we jump top to handle the others and by doing this, the speed problem fixed !!!

UPDATED

I just tried to make loop body smaller (in size (my first code)) and the result was amazing !!!!!

strl:
        push    rdi
        push    rsi
        mov     rdi, rcx
        mov     rsi, rcx
        mov     ecx, 0x00ff0000
        mov     edx, 0xff000000
        mov     r8, 0x000000ff00000000
        mov     r9, 0x0000ff0000000000
        mov     r10, 0x00ff000000000000
        mov     r11, 0xff00000000000000
        test    dil, 7
        jz      @f
        ; handle unaligned
  align 32
 @@:    mov     rax, QWORD [rdi]
        test    al, al
        jz      SHORT .end
        test    ah, ah
        jz      SHORT .end1
        test    eax, ecx
        jz      SHORT .end2
        test    eax, edx
        jz      SHORT .end3
        test    rax, r8
        jz      SHORT .end4
        test    rax, r9
        jz      SHORT .end5
        test    rax, r10
        jz      SHORT .end6
        test    rax, r11
        jz      SHORT .end7
        add     rdi, 8
        jmp     @b
  align 16
 .end:  mov     rax, rdi
        sub     rax, rsi
        pop     rsi
        pop     rdi
        ret
  align 16
 .end1: lea     rax, [rdi+1]
        sub     rax, rsi
        pop     rsi
        pop     rdi
        ret
  align 16
 .end2: lea     rax, [rdi+2]
        sub     rax, rsi
        pop     rsi
        pop     rdi
        ret
  align 16
 .end3: lea     rax, [rdi+3]
        sub     rax, rsi
        pop     rsi
        pop     rdi
        ret
  align 16
 .end4: lea     rax, [rdi+4]
        sub     rax, rsi
        pop     rsi
        pop     rdi
        ret
  align 16
 .end5: lea     rax, [rdi+5]
        sub     rax, rsi
        pop     rsi
        pop     rdi
        ret
  align 16
 .end6: lea     rax, [rdi+6]
        sub     rax, rsi
        pop     rsi
        pop     rdi
        ret
  align 16
 .end7: lea     rax, [rdi+7]
        sub     rax, rsi
        pop     rsi
        pop     rdi
        ret      

huangapple
  • 本文由 发表于 2023年5月24日 23:35:38
  • 转载请务必保留本文链接:https://go.coder-hub.com/76325226.html
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定