写一个函数，求一个字符串的长度，在main函数中输入字符串，并输出其长度...

固定值的加法。我用bx和ax作为寄存器。bx表示第一个固定值，ax表示第二个固定值，然后将其相加。

Proteus和labview仿真，有源码，非常不错的-Proteus and the LabVIEW simulation, source code has a very good

五个按键：关机off键，开机on键，五分钟定时键，十分钟定时键，二十分钟定时键，按相应的按键实现相应的功能，输出4HZ的方波-5 buttons: off off key, turn on key, five minutes from time to time key, 10 minutes from time to time key, 20 minutes from time to time, press the corresponding button to achieve the corresponding functions of the square wave output of 4HZ

汇编语言做的 俄罗斯方块 演示-compilation of the Russian language to block demonstrations

4相6线步进电机程序，用于测试步进电机，驱动采用反向驱动-4-phase stepper motor 6-line procedures for testing stepper motor, drive the use of reverse drive

一款打字游戏,有几分，暂停，调速等功能，界面优美，功能强大-a typing game

通过MIPS的汇编语言实现整型浮点数的加法运算。其中可以设计的指令有mfc1和mtc1。如输入1,1，则输出2.0，以此实现运算。

文件加密程序,汇编语言编写程序,源代码基本代码-File encryption procedures, assembly language programming, source code basic code

哲学家就餐问题是由Dijkatra提出并解决的典型同步问题。该问题描述的是有五个哲学家共用一张圆桌，分别坐在周围的五张椅子上，圆桌上有五个碗和五只筷子，他们的生活方式是交替的进行思考和就餐。平时，一个哲学家进行思考，饥饿时便试图取用其左右最靠近他的筷子，只有在他拿到两只筷子时才能进餐。进餐完，放下筷子继续思考。-Dining philosophers problem is presented and solved Dijkatra typical synchronization problems. Description of the problem is that there are five philosophers share a round-table were sitting around the five chair, round table has five bowl and five chopsticks, their way of life is the turn of thinking and eating. Normally, a philosopher to think, hunger when trying to access their nearest him about the chopsticks, only if he can get two chopsticks when eating. End meals, to lay down their chopsticks to continue thinking.

查看本地文件信息-View local file information