刚开始学汇编，完全一头雾水啊(*꒦ິ⌓꒦ີ)
求，救一救
data segment
tb db 10
tw dw 10
data ends
code segment
main proc far
assume cs:code,ds:data
start:
push ds
sub ax,ax
push ax
mov ax,data
mov ds,ax
mov ax,1234H
mov bx,4567H
push ax
push bx
pop ax
;1.SP=?,ax=?
pop bx
mov ax,-32767
add ax,-1
;2.ax=?(十六进制表示),SF=?,OF=?,CF=?
add ax,-1
;3.ax=?(十六进制及十进制表示),SF=?,OF=?,CF=?
mov ax,-5
cmp ax,-6
;4.ax=?(十六进制表示),SF=?,OF=?,CF=?
neg ax
cmp ax,6
;5.ax=?(十六进制表示),SF=?,OF=?,CF=?
mov ax,-10000
mov dx,0
mov bx,100
div bx
;6.ax=?,dx=?
mov ax,-10000
mov dx,0
mov bx,100
idiv bx
;7.ax=?,dx=?
mov ax,-10000
cwd
mov bx,100
idiv bx
;8.ax=?,dx=?
mov al,12h
and al,0fh
;9.al=?
mov al,-5
test al,80h
;10.zf=?
mov al,-5
sar al,1
;11.al=?
mov ax,1234h
mov cl,4
rol ax,cl
;12.ax=?
mov ax,1234h
push ax
call p1
pop ax
mov ax,-5
mov bx,6
cmp ax,bx
jb n1
;14.上条指令是否会跳转?
nop
n1:
cmp ax,bx
jl n2
;15.上条指令是否会跳转?
nop
n2:
ret
main endp
p1 proc near
push ax
mov ax,4567h
;13.请写出此时的堆栈内容
pop ax
ret
p1 endp
code ends
end