我必须想出一个 ASM 代码(对于 emu8086),它将在任何给定大小的数组中找到最小值和最大值。在示例代码中,我的讲师提供(看起来是)一个包含名为 LIST 的数组的数据段。他声称他会将这个列表替换为其他不同大小的列表,我们的代码必须能够处理它。
这是下面的示例代码。我已经突出显示了我添加的部分,只是为了向您展示我已尽力解决此问题:
; You may customize this and other start-up templates;
; The location of this template is c:\emu8086\inc\0_com_template.txt
org 100h
data segment
LIST DB 05H, 31H, 34H, 30H, 38H, 37H
MINIMUM DB ?
MAXIMUM DB ?
AVARAGE DB ?
**SIZE=$-OFFSET LIST**
ends
stack segment **;**
DW 128 DUP(0) **; I have NO CLUE what this is supposed to do**
ends **;**
code segment
start proc far
; set segment registers:
MOV AX,DATA **;**
MOV DS,AX **;I'm not sure what the point of this is, especially since I'm supposed to be the programmer, not my teacher.**
MOV ES,AX **;**
; add your code here
**;the number of elements in LIST is SIZE (I think)
MOV CX,SIZE ;a loop counter, I think
;find the minimum value in LIST and store it into MINIMUM
;begin loop
AGAIN1:
LEA SI,LIST
LEA DI,MINIMUM
MOV AL,[SI]
CMP AL,[SI+1]
If carry flag=1:{I got no idea}
LOOP AGAIN1
;find the maximum value in LIST and store it into MAXIMUM
;Something similar to the other loop, but this time I gotta find the max.
AGAIN2:
LEA SI,LIST
LEA DI,MINIMUM
MOV AL,[SI]
CMP AL,[SI-1] ;???
LOOP AGAIN2
**
; exit to operating system.
MOV AX,4C00H
INT 21H
start endp
ends
end start ; set entry point and stop the assembler.
ret