Compilation fill in the blank Thank you is simple, but I won t

Under the version (XP system) the debugging passes.

dseg segment

buf db 5,-4,0,3,100,-51cnt dw $-bufplus db "y=1",0ah,0dh,"$"

zero db "y=0",0ah,0dh,"$"

minu db "y=-1",0ah,0dh,"$"

dseg ends

cseg segment

assume cs:cseg,ds:dsegbegin: mov ax,dsegmov ds,ax

lea si,buf

mov cx,cnt

next: mov al,[si]and al,al

js fujz ling

lea dx,plus

jmp disp

fu: lea dx,minujmp disp

ling: lea dx,zerodisp: mov ah,9

int 21h

inc si

loop next

mov ah,4ch

int 21h

cseg ends

end begin

This is a program with a two-cycle structure.

Outer loop: controls the read character and determines whether the "e" determines whether the loop continues or ends;

Inner Circulation: The number of occurrences of "1" is recorded by shifting left and judging the state of the CF flag.

The procedure is as follows, and it has been debugged.

cseg segment

assume cs:cseg

begin:

mov ah,1

int 21h;Enter a character.

cmp al,'e'

jz stop;If it is "e", then go to stopxor dl, dl; Register zero for recording 1 number.

mov cx,8;The inner layer is cycled 8 times.

next: shl al,1;Shift 1 bit to the left, the highest is to move in cfjnc skip; If cf=0, go to skipinc dl; CF=1, DL increases by 1

skip: loop next;cx-1, if cx is not zero, then go to the next loop execution.

or dl,30h;Convert the number of occurrences of 1 to the corresponding ASCII code.

mov ah,2

int 21h;The number 2 function is called to display the number of times 1.

mov dl,0ah

int 21h

mov dl,0dh

int 21h;Show wrap carriage returns.

jmp begin;Go to Begin and read in the next character.

stop: mov ah,4ch

int 21h;End of program.

cseg ends

end begin

mov ax,var

cmp ax,0 ）

jz zero ;is equal to 0, then turn.

jg great；If it is greater than 0, it will be turned.

mov ax, 0ffffh；Less than 0

jmp short next

zero: mov ax,0

jmp next ）

great: mov ax,0001h

next: mov result,ax

① inc s10 ；The number of 100 is added to 1

jmp short change_addr ；short is a pseudo command, and the redirected label is change addr

loop compare ；Circulate.

data1 starts as the hexadecimal number corresponding to (256-35), 33h, 35h

data2 starts at the lower 16 digits of the hexadecimal number corresponding to (65536-35), the upper sixteen digits, 33h, 00,00,00,35h,00,00,00

data1 dd，35

data2 ffdd，0035

But in memory, the storage is high and low. That is, the high position is at the high address, and the status is at the low address.