在座的任何人都可以让我理解此程序吗,我将从中得到什么输出?

时间:2019-04-01 19:48:59

标签: assembly dos x86-16 emu8086

DATA SEGMENT
STRING1 DB 11H,22H,33H,44H,55H
MSG1    DB "FOUND$"
MSG2    DB "NOT FOUND$"
SE      DB 34H
DATA ENDS

PRINT MACRO MSG
    MOV AH, 09H
    LEA DX, MSG
    INT 21H
    INT 3
ENDM

CODE SEGMENT
ASSUME CS:CODE, DS:DATA
START:
    MOV AX, DATA
    MOV DS, AX
    MOV AL, SE
    LEA SI, STRING1
    MOV CX, 04H

UP:
    MOV BL,[SI]
    CMP AL, BL
    JZ FO
    INC SI
    DEC CX
    JNZ UP

    PRINT MSG2
    JMP END1
FO:
    PRINT MSG1
END1:
    INT 3

CODE ENDS
END START

3 个答案:

答案 0 :(得分:3)

程序在序列11H,22H,33H,44H中搜索字节'34H'。

START部分执行以下操作:

  • 将DS:SI设置为STRING1的地址。
  • 将AL设置为要搜索的字节 对于。
  • 将CX设置为要在STRING1(4 字节)。

循环部分执行以下操作:

  • 将DS:SI(从STRING1)处的字节加载到BL
  • 将其与要搜索的字节进行比较(在AL中)
  • 如果字节相等(零标志= 1),则打印“找到”
  • 否则(不相等)转到下一个字节(INC SI),递减计数器(CX)
  • 如果计数器为零,则退出循环并打印“找不到”

打印宏:

MOV AH, 09H
INT 21H 

是MS-DOS调用Print直到“ $” INT 3表示退出程序。

可疑的是,计数器CX设置为4,但STRING1序列包含5个字节。如果没有真正调试程序,我会说CX = 5是否正确?

答案 1 :(得分:0)

我的投票相当于memchr(s,c,n),其中sSI中(每次迭代递增),cBL中(与从SI获取“找到项目”退出条件的结果),nCX中(递减,如果未找到等于BL的数组元素,则为零,这是其他退出条件)加上友好的输出(包含int21h调用的宏)。参见http://man7.org/linux/man-pages/man3/memchr.3.html

答案 2 :(得分:0)

它在字节数组33h中搜索值11H,22H,33H,44H,55H

DATA SEGMENT
# Data where we will look up for value
STRING1 DB 11H,22H,33H,44H,55H
# message to be printed out
MSG1 DB "FOUND$"
# yet one message
MSG2 DB "NOT FOUND$"
# program will look up for this value
SE DB 33H

DATA ENDS

# this is PRINT macro- will be used later like a function
PRINT MACRO MSG
# 09H - is a function of interrupt 21h - print to console http://spike.scu.edu.au/~barry/interrupts.html
MOV AH, 09H
LEA DX, MSG
# print message using interrupt 21h
INT 21H
# I think it's exit - check it in the x86 interrupts table if you need
INT 3

ENDM # end of print macro

CODE SEGMENT

ASSUME CS:CODE, DS:DATA

START:

# this is the first line of code
MOV AX, DATA
MOV DS, AX
# copy value of SE (33h) to the AL register so later we can compare it
MOV AL, SE
LEA SI, STRING1

# save in CX length of the data to be looked up for value (actually num or iterations)
MOV CX, 04H

# the main loop starts here
UP:
# copy byte from buffer STRING1 to the BL register
MOV BL,[SI]
# check if AL==BL -> value found
CMP AL, BL

# if AL==BLjump to FO (will print FOUND)
JZ FO
# if not found, 
# move to next byte in STRING1
INC SI
# decrement loop constraint (loop should end before reaching end of string)
DEC CX
# if CX <> 0 loop again
JNZ UP
# if CX==0 print message NOT FOUND
PRINT MSG2
JMP END1
FO:
PRINT MSG1
END1:
INT 3
CODE ENDS

END START