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[assembly language] final1 DL : 0: 红绿灯控制 intel 8086cpu设计 8259 82-Intel 8086cpu traffic lights to control the design of 825,982
Date : 2026-01-08 Size : 2kb User :
[assembly language] 8086 DL : 0: 设有十个学生的成绩分别为56 69 84 82 73 88 99 63 100 80。编制程序分别统计低于60分，60-90 67-97，80-89，90-99和100分的人数，并存放于s5,s6,s7,s8,s9,s10中 -With 10 pupils, respectively 56 69 84 82 73 88 99 63 100 80. Statistical programming, respectively, below 60 60-90 67-97,80-89,90-99 and 100 the number of points, and deposited in the s5, s6, s7, s8, s9, s10 of
Date : 2026-01-08 Size : 125kb User : wjj
[assembly language] xt00 DL : 0: 用8086，8255，8259和8254构造系统实现对指示灯控制。 8255的PA0，PA1,PA2的三位DIP开关，通过DIP开关的闭合状态决定接在 PB口上的八个指示灯之一闪烁。如PA2,PA1,PA0为000时，PB0上所接的指示灯闪烁，其余灯熄灭。要求闪烁频率为每秒10次。设8259地址为20H和 21H，8255地址为60H～63H，8254地址为40H～43H,时钟频率为50KHz， 8259中断向量号为70H和71H.试设计硬件连接电路，填写中断向量表，编写全部初始化程序，完成控制程序编写。-8086,8255,8259 and 8254 by tectonic system of light control. 8255 of PA0, PA1, PA2 three DIP switches, DIP switches, through the closed state in the PB I decided to take on one of the eight indicator light flashes. Such as PA2, PA1, PA0 for 000 hours, PB0 on the indicator light flashes then the remaining light extinguish. Requirements flicker frequency of 10 times per second. 8259 address 20H located and 21H, 8255 address 60H ~ 63H, 8254 address 40H ~ 43H, the clock frequency of 50KHz, 8259 interrupt vector number is 70H and 71H. Try to design hardware to connect the circuit to fill in Interrupt Vector Table, the preparation of all the early the beginning of the procedure to complete the preparation of control procedures.
Date : 2026-01-08 Size : 3kb User : 林俊杰
[assembly language] (2) DL : 0: 用减奇数开平方运算 8086/8088指令系统中有乘除法指令但没有开平方指令，因此，开平方运算是通过程序来实现的。用减奇数法可求得近似平方根，获得平方根的整数部分。我们知道，N个自然数中的奇数之和等于N2 ，即： 1+3+5=32 1+3+5+7=16=42 1+3+5+7+9+11+13+15=64+82-Odd square root operation with reduced instruction set in 8086/8088 multiply and divide instructions, but no square root instruction, therefore, square root operation is achieved through the process. Reduction method can be obtained with approximate square root of the odd, the integer part to obtain the square root. We know, N a natural number equal to the sum of the odd N2, namely: 1+3+5 = 32 1+3+5+7 = 16 = 42 1+3+5+7+9+11+13+15 = 64+82
Date : 2026-01-08 Size : 4kb User : ada