利用8051串行口，和并行输出串行移位寄存器74LS164扩展一位数码显示在数码显示器上循环显示0...

用汇编编的匿名管道，支持win98/win2000/xp-Compilation made by the anonymous pipes, support win98/win2000/xp

8个发光管模拟二进制加法运算.asm 8个发光管模拟二进制加法运算.asm-8 LED simulated binary addition operations. Asm 8 months LED simulated binary addition operations. Asm

利用8253的秒表计时程序，中断的源代码-Procedures for the use of 8253" s stopwatch timing, interrupt source

输入三个整数x,y,z，请把这三个数由小到大输出。 1.程序分析：我们想办法把最小的数放到x上，先将x与y进行比较，如果x>y则将x与y的值进行交换，然后再用x与z进行比较，如果x>z则将x与z的值进行交换，这样能使x最小。-Enter the three integers x, y, z, please these three the number of small to large output. 1. Program analysis: We think of a way to put x number of the smallest, the first comparison of x and y, if x> y will be the value of x and y exchange and then to compare x and z, if x > z will be x and z values of the exchange, which will enable x minimum.

一个C style Assembler的source code-one of the C style Assembler source code

根据8051单片机的特点及交通灯在实际控制中的特点，本文提出一种用单片机自动控制交通灯及时间显示的方法.-According to the characteristics of 8051 and the traffic lights in the actual control of the characteristics of this paper, a single-chip automatic control with traffic lights and the time display method.

红外键值码LED显示,用于C51程序,解码遥控器编码 -IR key code LED display for the C51 process, decoding the remote control code

EDIT Assembler can for VC6 VC8 VC2005

PCI信息查询，汇编语言，基于8086系统，功能丰富-Electronic alarm clock, assembly language, based on 8086 system

学生成绩核算系统的设计与实现（25） 要求：（1）按班级按课程从文件中读入相应的平时成绩、期中考试成绩和期末考试成绩。 （2）三个成绩对总评成绩的百分比被定义为常数，各占总成绩的30 、30 和40 。 （3）计算每位学生的总评成绩。 （4）计算该班级本课程的总平均成绩。 （5）计算处于优、良、中、及格、不及格的学生人数以及占总人数的百分比。其中100-90为优，89-80为良，79-70为中，69-60为及格，60分以下为不及格。 （6）按要求输出成绩在优、良、中、及格、不及格各区间的学生学号、成绩。-The student achievement calculation system s design with realizes (25) Request: (1) reads in the corresponding usual result, the midterm examination result and the terminal examinations result according to the class and grade according to the curriculum from the document. (2) three result are defined to the general appraisal result s percentage as the constant, occupies the overall result respectively 30 , 30 and 40 . (3) calculates each student s general appraisal result. (4) calculates this class and grade this curriculum the population mean result. (5) calculates the student population which is in superiorly, good, passes an examination, does not pass an examination as well as accounts for the total number of peop