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win2000下的与下位机UART的串口通信程序
win2000下的与下位机UART的串口通信程序-WIN2000 with the crew under the UART serial communication program
- 2022-10-26 18:30:03下载
- 积分:1
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linux串口编程的一些文档以及一些示例,有助于刚开始接触串口的人的快速入门...
linux串口编程的一些文档以及一些示例,有助于刚开始接触串口的人的快速入门-some documentation, and sample some help contact the beginning of the serial Quick Start
- 2023-08-23 00:20:04下载
- 积分:1
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ATmega8单片机程序:直流电机PID控制器ATmega8,和VC + + + 6码RS232
Atmega8 Program: DC motor PID controller by atmega8, and VC+++6.0 code RS232
- 2022-05-25 06:10:56下载
- 积分:1
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具有语音自动应答功能的串口程序
具有语音自动应答功能的串口程序-with automatic voice response function of the Serial procedures
- 2022-05-22 15:02:23下载
- 积分:1
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VCL控件,功能强大,包括串口,Moden,传真
VCL控件,功能强大,包括串口,Moden,传真-VCL control, powerful, including serial, Moden, fax, etc.
- 2022-03-18 08:47:48下载
- 积分:1
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Communications routines 485 for them to learn from the reference. Communications...
485通讯例程,供大家学习参考。 -Communications routines 485 for them to learn from the reference. Communications routines 485 for them to learn from the reference.
- 2022-11-08 14:55:03下载
- 积分:1
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串口通信的东东 串口通信的东东串口通信的
串口通信的东东 串口通信的东东串口通信的-Dongdong serial communications serial communications serial communications Dongdong
- 2022-05-31 12:52:49下载
- 积分:1
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串行通信
串行通信-Serial Communication
- 2022-02-20 11:01:29下载
- 积分:1
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这是非常有挑战性的题目。对于用户输入的任意一个平面函数f(x),绘制出其函数曲线。这里最关键的技术难点就是如何实现计算表达式的值。在《编译原理》和《数据结构》的...
这是非常有挑战性的题目。对于用户输入的任意一个平面函数f(x),绘制出其函数曲线。这里最关键的技术难点就是如何实现计算表达式的值。在《编译原理》和《数据结构》的书中,都有对表达式运算方法的论述。说实在的,在编译型计算机语言中实现对用户输入表达式的运算是非常困难的。需要对表达式进行扫描,去括号,按照运算符的优先级生成2叉树,然后遍历该树生成逆波兰表达式,再然后通过栈的方法进行运算。如果在表达式中再包含有函数的话......描述起来都麻烦,更不要说用程序实现了:-(-This is a very challenging issue. For user input an arbitrary planar function f (x), mapping out its function curves. Here the most critical technical problems is how to achieve calculated the value of the expression. "Compiling Principle" and "data structure," the book, have right of expression Operational discussed. Indeed, the computer language compiler to achieve expression of the user input operation is very difficult. Need to scan expression to the brackets, in accordance with the priority Operators Generation 2-ary tree, and then traverse the tree generated reverse Polish expression, and then through the stack approach to computing. If the expression again contains the function described ...... these are trouble, not to mention t
- 2022-02-25 13:48:40下载
- 积分:1
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一个串口测试程序,包括源代码,供大家学习
一个串口测试程序,包括源代码,供大家学习-A serial test procedure, including the source code for the U.S. study
- 2022-02-28 13:16:20下载
- 积分:1