[Original] demonstrates the compilation source program for functions that determine whether a string is a substring of another string

; @@@@@@@@@@@@
Function: shows a function to determine whether a string is a substring of another string.
; Author: Huang zhibin at Hechi, Guangxi
; Capacity within the daily period
;----------------------------------------------------
; 2002.06.25 complete basic functions
; 2002.06.26 modify issubstr to reduce one byte
. An error occurred while modifying the algorithm used on the 26th. Modify the calculation string 1.
In string 2, the file length is reduced to 320 bytes.

Codeseg segment
; ========================================================== ================
Org 100 h
Assume ES: codeseg, SS: codeseg, CS: codeseg, DS: codeseg
;-----------------------------
Main proc
MoV dx, offset strprompt
MoV ah, 9
Int 21 h
MoV dx, offset str1maxlen
MoV ah, 0ah
Int 21 h

Call crline

MoV cstrname, '2'
MoV dx, offset strprompt
MoV ah, 9
Int 21 h
MoV dx, offset str2maxlen
MoV ah, 0ah
Int 21 h

Call crline

XOR ch, ch
MoV Cl, str1len
Xor bh, BH
MoV BL, str2len
MoV Si, offset string1
MoV Di, offset string2
Call issubstr
MoV dx, offset strisnull
CMP bp,-3
Je @ quit
MoV dx, offset strtoolen
CMP bp,-2
Je @ quit
MoV dx, offset strno
CMP bp,-1
Je @ quit
MoV Di, offset cposition
MoV ax, BP
Call dec2ascii
MoV dx, offset stryes
@ Quit:
MoV ah, 9
Int 21 h
MoV ax, 4c00h
Int 21 h
Main endp
;----------------------
Conststrlen equ 15
Strprompt DB "input no"
Cstrname dB '1'
DB "strings: $"
Str1maxlen dB conststrlen
Str1len db 0
String1 dB conststrlen DUP (?)
Str2maxlen dB conststrlen
Str2len db 0
String2 dB conststrlen DUP (?)
Strtoolen DB "string 1 is longer than string 2! $"
Strisnull DB "the length of string 1 is 0 $"
Stryes DB "string 1 appears on the left of string 2"
Cposition DB "location $"
Strno dB 0dh, 0ah, "string 1 is not a substring of string 2 $"

; ========================================================== ====
Function: convert data in ax to a corresponding decimal number string
; Input:
; AX = data to be converted
; DI = the first buffer address for storing the converted string
; Output: None
;--------------------------------------------
Dec2ascii proc
MoV CX, 2
MoV DL and 100
@ Loopdiv:
Div DL
Add Al, '0'
MoV [di], Al
INC di
MoV Al, ah
XOR ah, ah
Shr dl, 1
Loop @ loopdiv
Add Al, '0'
MoV [di], Al
RET
Dec2ascii endp

============================================
Function: determines whether string 1 is a substring of string 2.
; Entry:
; Cx = length of string 1
; BX = string 2 Length
; SI = start address of string 1
; DI = start address of string 2
Export:
; BP> = 0 string 1 is the substring of string 2, and BP = the position where string 1 first appears in string 2 (starting from 0)
; BP =-1 string 1 is not a substring of string 2
; BP =-2 string 1 is longer than string 2
; BP =-3 The length of string 1 is 0
;----------------------------------
Issubstr proc
Push di; Save the first address of string 2 for future computation of the position where string 1 appears in string 2
MoV bp,-3
Jcxz @ errresult
MoV bp,-2
Cmp cx, BX
JG @ errresult
ClD
@ Loopcmp:
Push CX
Push Si
Push di
Repe cmpsb
Je @ yesresult
Pop di; restore string 2 the first address of this comparison
INC di; adds 1 to the first address of string 2 to point to the next character for next comparison
Dec BX; string 2 length minus 1
Pop Si; restore the first address of string 1
Pop CX; restore string 1 Length
Cmp bx, CX
Jge @ loopcmp
MoV bp,-1
@ Errresult:
Pop CX
RET
@ Yesresult:
The following method calculates the position of string 1 in string 2.
Pop ax; these two items are the Di, Si
; Pop ax
; Pop CX; send string 1 to CX
; Pop bp; send the first address of the original string to BP
; Sub Di, CX; Calculate the first address of string 2 in this comparison
; Sub Di, BP; position of string 1 in string 2 = first address of string 2 in this comparison-first address of original string 2
; MoV bp, Di
; RET
Pop bp; pop the previously pushed Di, that is, the first address of string 2 in this comparison.
Pop ax; pop up Si, useless
Pop CX; send string 1 to CX
Pop di; send the first address of the original string 2 to Di
Sub bp, di; Calculate the position of string 1 in string 2
RET
Issubstr endp

====================================
Crline proc
MoV dx, offset strcrline
MoV ah, 9
Int 21 h
RET
Crline endp
;----------------------
Strcrline dB 0dh, 0ah, '$'
Codeseg ends
End main

; Error algorithm:
Because the address of string 2 is not restored, when string 1 is "123" and string 2 is "12345"
The running result is "string 1 is not a substring of string 2"
Issubstr proc
Push di; save string 2 length for later calculation of the position where string 1 appears in string 2
MoV bp,-3
Jcxz @ errresult
MoV bp,-2
Cmp cx, BX
JG @ errresult
ClD
@ Loopcmp:
Push CX
Push Si
Repe cmpsb
Je @ yesresult; jump if it is a substring
Pop Si

The following three commands are the latest remaining length of string 2.
; Number of bytes compared by the system = string 1-CX
; BX = string 2 Original Residual Length
; String 2 latest remaining length = string 2 Original remaining length-number of bytes that have been compared by the System
; = Bx-(string 1-cx)
; = Bx + cx-string 1 Length

Add Bx, CX; before execution BX = string 2 original length
Pop CX; restore the first address of string 1
Sub BX, CX; Cx = string 1 length, after execution BX = string 2 latest remaining length
Cmp bx, CX
Jge @ loopcmp
MoV bp,-1
@ Errresult:
Pop CX
RET
@ Yesresult:
Pop ax; this is the Si
Pop CX; send string 1 to CX
Pop bp; send string 2 to BP
Sub Di, CX; the following three commands calculate the position of string 1 in string 2 stored in BP
Sub Di, BP
MoV bp, Di
RET
Issubstr endp
; ========================================================== ====
Codeseg ends
End main