Assembly 下面的汇编语言代码是如何运行的?
我对组装一无所知,但我被分配了一项任务 请告诉我下面的代码是如何运行的?我指的是步骤或程序Assembly 下面的汇编语言代码是如何运行的?,assembly,x86,masm,irvine32,Assembly,X86,Masm,Irvine32,我对组装一无所知,但我被分配了一项任务 请告诉我下面的代码是如何运行的?我指的是步骤或程序 TITLE MASM Template (main.asm) ; Description: ; ; Revision date:f INCLUDE Irvine32.inc .data counter dword 1; instruct1 BYTE"How many line are required?: ",0 ;give instruction to user to give the input
TITLE MASM Template (main.asm)
; Description:
;
; Revision date:f
INCLUDE Irvine32.inc
.data
counter dword 1;
instruct1 BYTE"How many line are required?: ",0 ;give instruction to user to give the input
answer BYTE"Answer:",0
newline BYTE 0Dh, 0Ah
sum BYTE 0
.code
main PROC
mov edx,OFFSET instruct1 ;move instruct1 to edx
call WriteString
call readint;
mov ecx,eax; move ecx to eax
L1:
push ecx;
mov ecx,counter
L2:
mov al,'*';add '*' into al
call writechar;
loop l2;
pop ecx;
inc counter;
call crlf;
loop l1;
exit
main ENDP
end main
此代码打印提示并输入数字。然后它会打印出恒星的行数。第一行是1颗星,第二行是2颗星,依此类推。我对代码进行了注释,以使其更加清晰 代码使用两个嵌套循环来实现这一点。
ecx
寄存器用于两个循环:作为每行上星号的计数器,以及行计数。这就是为什么会按下并弹出ecx
,这样它就可以在内部循环中有另一个计数
TITLE MASM Template (main.asm) ;used for listings etc.
; Description:
;
; Revision date:f
INCLUDE Irvine32.inc ;include another code file
.data ;data segment
counter dword 1 ;characters per line
instruct1 BYTE"How many line are required?: ",0
answer BYTE"Answer:",0 ;irrelevant to this code
newline BYTE 0Dh, 0Ah ;used by crlf
sum BYTE 0 ;irrelevant to this code
.code ;code segment
main PROC ;declare code block
mov edx,OFFSET instruct1 ;message pointer
call WriteString ;display message
call readint ;input an integer
mov ecx,eax ;move input to line loop register
L1:
push ecx ;save line count register
mov ecx,counter ;load character counter
L2:
mov al,'*' ;the char we wish to print
call writechar ;output one char
loop L2 ;next character (count in cx)
pop ecx ;restore the line counter
inc counter ;increment characters per line
call crlf ;print the newline defined above
loop L1 ;next line (count in cx)
exit ;return to OS
main ENDP ;end code block
end main ;end of code file
如果输入为3,则输出为:
*
**
***
顺便说一句,我会批评代码的作者,原因有二
mov ecx,eax ; move ecx to eax
理由一。评论是背对背的;将行计数器的返回值eax
移动到ecx
理由2:永远不要用注释来解释指令的作用,你可以用RTM来解释。使用注释来增加价值,以明确其目的。猜猜
调用WriteString
和调用readint
做什么,然后查看字符串和注释,也许可以从那里开始工作?你能解释一下调用WriteString
和调用readint
是什么吗?tq它所做的是打印一个正方形的*
字符。因此,如果用户输入2,它将打印两行**
。如果用户输入4,它将打印4行***
。如果你问“它是如何工作的?”,你必须更具体。WriteString
和ReadInt
是包含在文件顶部的包含文件“Irvine32.inc”中最有可能提供的函数。我怀疑WriteString
向控制台写入字符串,而ReadInt
读取用户从控制台输入的数字。