Help, programming in assembly language

org 0000h

ajmp main

org 0030h

main:mov 31h,#

mov r0,#40h ;R0 is used as an on-chip RAM data pointer, pointing to 40HMOV R2, 16; R2 was used as a counter, and the cycle was 16 times.

clr a ;a is used to find the sum of accumulations and is initialized to 0

s0:add a,@r0

jnc next ;After adding, whether there is a carry or not, and if there is a carry added to the upper 8 digits.

push acc ;Because MCS51 addition can only use A as the destination operand, the value in A must be protected first.

mov a,31h

addc a,#0 ;Indicates a 0+c > AMOV 31H,A

pop acc

next:inc r0 ;Modify the data pointer to point to the next cell.

djnz r2,s0 ;r2-1 > whether r2 is 0, if it is not 0, it is looped.

mov 30h,a

sjmp $end

Set data1=40h data2=0h data3=0xunhuan:

mov a,data1

mov mp1,a write data1 to mp1 (addressing pointer) mov a,[02] write the data of the addressing pointer address to accadd a,data2 data2+acc=accmov data2,a acc data2mov a,0h 0 acc

adc a,data3 data3+c+acc=accmov data3,a acc data3inc data1 increment data1

mov a,50h

xor a,data1

SNZ z data1 is equal to 50h and skips unequal to continue the cycle.

jmp xunhun

mov a,30h

mov mp1,a

mov a,data2

mov [02],a put data2 [30h]inc mp1

mov a,data3

mov [02],a put data3 [31h]end

The answer is guess] :d

Assembly language is the low-level language of the wild staring module for computers, and assembly language source programs are programs written in assembly language. Therefore, choose D.

Pick D. Pay attention to the lower priority, the priority of AND is higher than that of OR and XOR 00000110 (6).

with 00000010(2).

or 00001100(12).

00001110（0eh）

0eh is greater than or equal to 0eh, and the result is true.

In the assembly, all 1s are used, and num1 is defined as a byte type, so it is 8 1s - 0ffh.

In the same way, 6 and 2 = 2, 12 xor 2 = 0eh, 0eh le 0eh=0ffh.