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196 lines
4.3 KiB
196 lines
4.3 KiB
page ,132
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title memcmp - compare to blocks of memory
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;***
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;memcmp.asm - compare two blocks of memory
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;
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; Copyright (c) 1985-2001, Microsoft Corporation. All rights reserved.
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;
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;Purpose:
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; defines memcmp() - compare two memory blocks lexically and
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; find their order.
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;
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;Revision History:
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; 05-16-83 RN initial version
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; 07-20-87 SKS rewritten for speed
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; 05-17-88 SJM Add model-independent (large model) ifdef
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; 08-04-88 SJM convert to cruntime/ add 32-bit support
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; 08-23-88 JCR 386 cleanup
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; 10-25-88 JCR General cleanup for 386-only code
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; 03-23-90 GJF Changed to _stdcall. Also, fixed the copyright.
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; 05-10-91 GJF Back to _cdecl, sigh...
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; 05-01-95 GJF New, faster version from Intel!
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; 05-09-95 GJF Replaced a bad ret with a good jz retnull (my editing
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; error, not Intel's).
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; 11-06-95 GJF Corrected order in which esi and edi are saved.
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;
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;*******************************************************************************
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.xlist
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include cruntime.inc
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.list
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page
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;***
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;int memcmp(buf1, buf2, count) - compare memory for lexical order
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;
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;Purpose:
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; Compares count bytes of memory starting at buf1 and buf2
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; and find if equal or which one is first in lexical order.
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;
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; Algorithm:
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; int
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; memcmp (buf1, buf2, count)
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; char *buf1, *buf2;
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; unsigned count;
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; {
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; if (!count)
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; return(0);
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; while (--count && *buf1 == *buf2)
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; {
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; buf1++;
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; buf2++;
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; }
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; return(*buf1 - *buf2);
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; }
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;
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;Entry:
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; char *buf1, *buf2 - pointers to memory sections to compare
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; unsigned count - length of sections to compare
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;
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;Exit:
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; returns -1 if buf1 < buf2
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; returns 0 if buf1 == buf2
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; returns +1 if buf1 > buf2
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;
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;Uses:
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;
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;Exceptions:
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;
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;*******************************************************************************
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CODESEG
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public memcmp
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memcmp proc
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.FPO ( 0, 3, 0, 0, 0, 0 )
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mov eax,[esp+0ch] ; eax = counter
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test eax,eax ; test if counter is zero
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jz short retnull ; return 0
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mov edx,[esp+4] ; edx = buf1
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push esi
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push edi
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mov esi,edx ; esi = buf1
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mov edi,[esp+10h] ; edi = buf2
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; Check for dword (32 bit) alignment
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or edx,edi
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and edx,3 ; edx=0 iff buf1 are buf2 are aligned
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jz short dwords
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; Strings are not aligned. If the caller knows the strings (buf1 and buf2) are
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; different, the function may be called with length like -1. The difference
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; may be found in the last dword of aligned string, and because the other
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; string is misaligned it may cause page fault. So, to be safe. the comparison
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; must be done byte by byte.
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test eax,1
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jz short main_loop
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mov cl,[esi]
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cmp cl,[edi]
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jne short not_equal
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inc esi
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inc edi
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dec eax
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jz short done ; eax is already 0
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main_loop:
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mov cl,[esi]
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mov dl,[edi]
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cmp cl,dl
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jne short not_equal
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mov cl,[esi+1]
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mov dl,[edi+1]
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cmp cl,dl
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jne short not_equal
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add edi,2
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add esi,2
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sub eax,2
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jnz short main_loop
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done:
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pop edi
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pop esi
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retnull:
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ret ; _cdecl return
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dwords:
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mov ecx,eax
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and eax,3 ; eax= counter for tail loop
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shr ecx,2
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jz short tail_loop_start
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; counter was >=4 so may check one dword
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rep cmpsd
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jz short tail_loop_start
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; in last dword was difference
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mov ecx,[esi-4] ; load last dword from buf1 to ecx
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mov edx,[edi-4] ; load last dword from buf2 to edx
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cmp cl,dl ; test first bytes
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jne short difference_in_tail
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cmp ch,dh ; test seconds bytes
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jne short difference_in_tail
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shr ecx,10h
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shr edx,10h
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cmp cl,dl ; test third bytes
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jne short difference_in_tail
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cmp ch,dh ; they are different, but each one is bigger?
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; jmp short difference_in_tail
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difference_in_tail:
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mov eax,0
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; buf1 < buf2 buf1 > buf2
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not_equal:
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sbb eax,eax ; AX=-1, CY=1 AX=0, CY=0
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pop edi ; counter
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sbb eax,-1 ; AX=-1 AX=1
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pop esi
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ret ; _cdecl return
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; in tail loop we test last three bytes (esi and edi are aligned on dword
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; boundary)
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tail_loop_start:
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test eax,eax ; eax is counter%4 (number of bytes for tail
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; loop)
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jz short done ; taken if there is no tail bytes
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mov edx,[esi] ; load dword from buf1
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mov ecx,[edi] ; load dword from buf2
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cmp dl,cl ; test first bytes
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jne short difference_in_tail
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dec eax ; counter--
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jz short tail_done
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cmp dh,ch ; test second bytes
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jne short difference_in_tail
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dec eax ; counter--
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jz short tail_done
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and ecx,00ff0000h ; test third bytes
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and edx,00ff0000h
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cmp edx,ecx
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jne short difference_in_tail
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dec eax
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tail_done:
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pop edi
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pop esi
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ret ; _cdecl return
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memcmp endp
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end
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