Arrays 如何在NASM中实现2D阵列
我试图弄清楚如何打印出数组的行和列。该程序询问有多少行和列,并根据用户输入显示为[0][0]输入数字为[0][1]输入数字等 以下是我到目前为止所写的内容:Arrays 如何在NASM中实现2D阵列,arrays,assembly,nasm,sasm,Arrays,Assembly,Nasm,Sasm,我试图弄清楚如何打印出数组的行和列。该程序询问有多少行和列,并根据用户输入显示为[0][0]输入数字为[0][1]输入数字等 以下是我到目前为止所写的内容: %include "io.inc" SECTION .data ; Data section, initialized variables num_rows db "How many rows?: ",0 num_col db "How many columns?: ",0 prompt db "Enter a
%include "io.inc"
SECTION .data ; Data section, initialized variables
num_rows db "How many rows?: ",0
num_col db "How many columns?: ",0
prompt db "Enter a number for [%d][%d]:",10,0
sum db "The sum is: ",10,0
number db "%d",10,0
rows times 4 dd "%d",0
col times 4 dd "%d",0
arrayLen dd 9 ; length of array
;size equ rows*col
formatin db "%d", 0
section .bss
array resd 6; this is a test array for testing purposes
SECTION .text ; Code section.
global CMAIN ; the standard gcc entry point
extern printf ,scanf
CMAIN: ; the program label for the entry point
;-----Ask for the number of rows and display
push num_rows
call printf
add esp,4 ;remove the parameter
push rows ;address of rows
push formatin ;arguments are right to left
call scanf
add esp,8
;move the values into the registers
mov ebp,[rows]
push ebp
push number
call printf
add esp,8
;----Ask for the number of cols and display
push num_col
call printf
add esp,4
push col
push formatin
call scanf
add esp,8
;move the values into the registers
mov ebx, [col]
push ebx
push number
call printf
add esp,8
mov ebp,array
push ecx
push number
call printf
add esp,8
mov ecx,0
xor ebp,ebp
outerLoop:
mov edx,ecx
push ecx
mov ecx,0
inner:
push ecx
;output
push ecx
push edx
push prompt
call printf
add esp,12
;Get addr
push ecx
push edx
push esi
;call GetElement
add esp,12
;input
push eax
push number
call scanf
add esp,8
pop ecx
inc ecx
cmp ecx,[col]
jl inner
pop ecx
end_outer:
inc ecx
cmp ecx,[rows]
jl outerLoop
push sum
call printf
add esp,4
xor ebp,ebp <- My professor told me never to use this
ret
;GetElement: THIS WHOLE SUBPROGRAM IS COMMENTED
; mov ebx,[ebp+8] ;addr of array
; mov ecx,[ebp+12] ;row
; mov esi,[ebp+16] ;col
; mov eax,ecx
; mul dword [col]
; add eax,esi
; imul eax,4
; add eax,ebx
; leave
; ret
当我运行代码时,索引[rows][cols]无法正确打印。有人能指导我吗?你的问题与访问数组元素无关-你的逻辑是正确的,尽管imul eax,4是个坏主意,应该用shl eax,2或lea eax,[eax*4]或lea eax,[ebx+eax*4]代替,因为数组中的偏移量不是有符号的值,避免这种乘法更快 相反;你的问题是C调用约定很糟糕。它们用大量额外的指令来操作堆栈,使代码更难读取和调试,从而污染了代码;并对其进行优化,例如使用子esp。。。要为要传递给任何子函数和mov[rsp+…]的最大参数保留空间。。。而不是推。。。在调用子函数之前设置参数是痛苦的;整个过程最终都是一个容易出错且缓慢的混乱状态,这对于汇编来说是不必要的,除非您调用的是由C编译器编译的函数 更具体地说;对于GetElement,您使用ebp作为堆栈帧,但没有将ebp设置为堆栈帧,因此当函数尝试将堆栈中的参数获取到寄存器中时,函数不会从正确的位置获取参数 要真正遵守C调用约定CDECL,它应该更像:
GetElement:
push ebp
mov ebp,esp ;Set up ebp as stack frame
push ebx ;ebx is "callee preserved" (not "caller saved") and is modified, so it must be saved/restored
push esi ;esi is "callee preserved" (not "caller saved") and is modified, so it must be saved/restored
mov ebx,[ebp+3*4+4] ;addr of array
mov ecx,[ebp+3*4+4+4] ;row
mov esi,[ebp+3*4+4+8] ;col
mov eax,ecx
mul dword [col]
add eax,esi
lea eax,[ebx+eax*4]
pop esi
pop ebx
leave
ret
具有讽刺意味的是,对于您的代码,参数已经在寄存器中-唯一的调用方正在执行以下操作:
push ecx ;col
push edx ;row
push esi ;address of array
call GetElement
add esp,12
…这意味着,如果您忘记了C调用约定,您的GetElement可能会像这样删除10条不必要的指令:
;Inputs:
; ecx = column
; edx = row
; esi = address of array
;
;Outputs:
; eax = address of element in the array
;
;Trashed:
; edx
GetElement:
mov eax,edx
mul dword [col]
add eax,ecx
lea eax,[esi+eax*4]
ret
call GetElement
..调用代码可能如下所示:删除4条不必要的指令:
;Inputs:
; ecx = column
; edx = row
; esi = address of array
;
;Outputs:
; eax = address of element in the array
;
;Trashed:
; edx
GetElement:
mov eax,edx
mul dword [col]
add eax,ecx
lea eax,[esi+eax*4]
ret
call GetElement
如果从moveax、[col]开始,GetElement可能会更高效;imul eax,edx。无需使用速度较慢的mul,该mul需要一个或两个额外的uop将高半部写入EDX。另外,你也不会丢弃任何寄存器。