在 tasm 汇编语言中查找数组的索引并打印它 [英] finding index of the array in tasm assembly language and printing it

问题描述

我做了一个tasm汇编语言程序，它在用户输入的数组中找到最小值.我想找到程序正在查找的最小值的元素的索引.

I have made a tasm assembly language program, which finds the minimum in the user-inputted array. I want to find the index of the element of the minimum value which the program is finding.

我想找到程序找到的元素的索引.

I want to find the index of the element which the program finds.

例如:输入数组是[1,2,3,4,5,6].它应该返回 1 作为最小值，0 作为索引.

For example: input array is [1,2,3,4,5,6]. It should return 1 as minimum value and 0 as index.

这是代码.

Data Segment
 msg db 0dh,0ah,"Please enter the length of the array: $"
 msg1 db 0dh,0ah,"Enter a number: $"
 newl db 0dh,0ah," $"
 res db 0dh,0ah,"The minimum is: $"
 len db ?
 min db ?
Data ends
Code Segment
assume CS:Code,DS:Data
Start:
 mov ax,Data
 mov DS,ax

 mov dx,offset msg
 mov ah,09h
 int 21h
 
 call Accept

 mov len,bl
 mov cl,bl
 mov ch,00h

 mov di,1000h
 
back: mov dx,offset msg1
 mov ah,09h
 int 21h
 call Accept
 mov [di],bl
 inc di
 loop back

 mov di,1000h
 mov cl,len
 mov ch,00h

 mov dx,offset newl
 mov ah,09h
 int 21h

 mov al,[di] 
 mov min,al

chk: mov bl,min
 mov al,[di]
 cmp bl,al
 jc a
 mov min,al
 jmp b
a: mov min,bl
b: inc di
 loop chk

 mov dx,offset res
 mov ah,09h
 int 21h

 mov bl,min
 call DispNum

 mov ah,4ch
 int 21h
Accept proc
 mov ah,01h
 int 21h
 call AsciiToHex
 rol al,4
 mov bl,al
 mov ah,01h
 int 21h
 call AsciiToHex
 add bl,al
 ret
endp
DispNum proc
 mov dl,bl
 and dl,0f0h
 ror dl,4
 call HexToAscii
 mov ah,02h
 int 21h
 mov dl,bl
 and dl,0fh
 call HexToAscii
 mov ah,02h
 int 21h
endp
AsciiToHex proc
 cmp al,41h
 jc sk
 sub al,07h
sk: sub al,30h
 ret
endp
HexToAscii proc
 cmp dl,0ah
 jc sk2
 add dl,07h
sk2: add dl,30h
 ret
endp
Code ends
end Start

推荐答案

mov di,1000h

因为在您的 2 个十六进制数字输入例程中，用户可以输入的最大值是FF".(255)，您可以在程序的数据部分保留一个该大小的缓冲区，而不是仅仅相信偏移地址 1000h 不会与内存中的任何重要内容重叠.

Because in your 2 hexdigits input routine the biggest value that the user can type is "FF" (255), you could reserve a buffer of that size in your program's data section instead of just trusting that the offset address 1000h will not overlap with anything important in memory.

Data Segment
 buf db 256 dup (0)
 msg db 0dh,0ah,"Please enter the length of the array: $"
 ...

为了解决获取最小值所在数组元素的索引的任务，可以将DI中地址的当前值保存在一个额外的寄存器中，如SI 并在每次代码偶然发现较小的值时更新:

To tackle the task of getting the index of the array element where the minimum is located, you can save the current value of the address in DI in an extra register like SI and update that each time the code stumbles upon a smaller value:

 ...
 mov di, offset buf
 mov al, [di] 
 mov min, al
 mov si, di    ; Address of the first chosen minimum
chk:
 mov al, [di]
 cmp al, min
 jae a
 mov min, al   ; Smaller than anything before
 mov si, di    ; Remember the address of the newest minimum
a:
 inc di
 loop chk

在此代码之后，您要查找的索引是通过从找到最小值的地址中减去数组的开头获得的:

After this code the index that you're looking for is obtained from subtracting the start of the array from the address of where the minimum was found:

sub si, offset buf  ; The index of the minimum

---

通过引入 SI 地址带来了不再需要单独的 min 变量的机会:

---

With introducing that SI address comes the opportunity of no longer needing the separate min variable:

 ...
 mov di, offset buf
 mov si, di    ; Address of the first chosen minimum
chk:
 mov al, [di]
 cmp al, [si]  ; At [si] is the current minimum
 jae a
 mov si, di    ; Remember the address of the newest minimum
a:
 inc di
 dec cx
 jnz chk
 ;;; mov al, [si]    ; The minimum should you need it
 sub si, offset buf  ; The index of the minimum

请注意，我删除了较慢的 loop chk 指令并将其替换为 dec cx jnz chk 对.

Please note that I've removed the slower loop chk instruction and replaced it by the pair dec cx jnz chk.

这里 是一个非常非常相似的答案，它涉及最大值而不是最小值......

Here is a very, very similar answer that deals with the maximum instead of the minimum...

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