电子发烧友网：本资料是关于单片机及接口技术这门课程的期末考试试卷及答案的详解。 8.当需要从MCS-51单片机程序存储器取数据时，采用的指令为( )。 a)MOV A, @R1 b)MOVC A, @A + DPTR c)MOVX A, @ R0 d)MOVX A, @ DPTR 二、填空题(每空1分，共30分) 1.一个完整的微机系统由 和 两大部分组成。 2.8051 的引脚RST是____(IN脚还是OUT脚)，当其端出现____电平时,8051进入复位状态。8051一直维持这个值，直到RST脚收到____电平，8051才脱离复位状态，进入程序运行状态，从ROM H单元开始取指令并翻译和执行。 3.半导体存储器分成两大类 和 ，其中 具有易失性，常用于存储 。
上传时间： 2015-01-03
上传用户：wfl_yy
温湿度传感器 sht11 仿真程序 sbit out =P3^0; //加热口 //sbit input =P1^1;//检测口 //sbit speek =P2^0;//报警 sbit clo =P3^7;//时钟 sbit ST =P3^5;//开始 sbit EOC =P3^6;//成功信号 sbit gwei =P3^4;//个位 sbit swei =P3^3;//十位 sbit bwei =P3^2;//百位 sbit qwei =P3^1;//千位 sbit speak =P0^0;//报警音 sbit bjled =P0^1;//报警灯 sbit zcled =P0^2;//正常LED int count; uchar xianzhi;//取转换结果 uchar seth;//高时间 uchar setl;//低时间 uchar seth_mi;//高时间 uchar setl_mi;//低时间 bit hlbz;//高低标志 bit clbz; bit spbz; ///定时中断程序/// void t0 (void) interrupt 1 using 0 { TH0=(65536-200)/256;//5ms*200=1000ms=1s TL0=(65536-200)%256; clo=!clo;//产生时钟 if(count>5000) { if(hlbz) { if(seth_mi==0){seth_mi=seth;hlbz=0;out=0;} else seth_mi--; } if(!hlbz) { if(setl_mi==0){setl_mi=setl;hlbz=1;out=1;} else setl_mi--; } count=0; } else count++; } ///////////// ///////延时/////// delay(int i) { while(--i); } ///////显示处理/////// xianshi() { int abcd=0; int i; for (i=0;i<5;i++) { abcd=xianzhi; gwei=1; swei=1; bwei=1; qwei=1; P1=dispcode[abcd/1000]; qwei=0; delay(70); qwei=1; abcd=abcd%1000; P1=dispcode[abcd/100]; bwei=0; delay(70); bwei=1; abcd=abcd%100; P1=dispcode[abcd/10]; swei=0; delay(70); swei=1; abcd=abcd%10; P1=dispcode[abcd]; gwei=0; delay(70); gwei=1; } } doing() { if(xianzhi>100) {bjled=0;speak=1;zcled=1;} else {bjled=1;speak=0;zcled=0;} } void main(void) { seth=60;//h60秒 setl=90;//l90秒 seth_mi=60;//h60秒 setl_mi=90;//l90秒 TMOD=0X01;//定时0 16位工作模式 TH0=(65536-200)/256; TL0=(65536-200)%256; TR0=1; //开始计时 ET0=1; //开定时0中断 EA=1; //开全中断 while(1) { ST=0; _nop_(); ST=1; _nop_(); ST=0; // EOC=0; xianshi(); while(!EOC) { xianshi(); } xianzhi=P2; xianshi(); doing(); } }
上传时间： 2013-10-16
上传用户：黄蛋的蛋黄
Displaying a large bitmap file on a dialog box, in its original size, is quite difficult in the VC++ environment. However, it is possible to display a large bitmap to a predefined area of the dialog by using the StretchBlt( ) function.The major disadvantage of this is that the clarity of the image will be lost. Check out this article for displaying large bitmaps into the desired area of your dialog box in its original size with a scrolling technique used to show the entire bitmap. 滚动显示位图 在VC++环境下，在一个对话框中显示一个原始尺寸的大小的位图文件相当是困难的。然而,通过使用 StretchBlt()函数一个给定的区域显示一个大的位图是可能的。主要的缺点是图像将会失真。看了这篇通过卷动技术显示整个位图技术的文章，你将能够以它的原始尺寸在给定对话框的区域内显示一个大位图。 来源: http://www.codeguru.com/bitmap/ScrollBitmap.html
标签： Displaying difficult original bitmap
上传时间： 2014-01-05
上传用户：yiwen213
Mantis an easily deployable, web based bugtracker to aid product bug tracking. It requires PHP, MySQL and a web server. It is simpler than Bugzilla and easily editable. Check out the online Demo.
标签： deployable bugtracker requires tracking
上传时间： 2014-01-25
上传用户：小宝爱考拉
Setting up an ADOCE project using Visual C++ 6.0 is rather simple. Assuming that you have downloaded and installed the ADOCE SDK from Microsoft, you are ready to use it in your Windows CE Database applications. The sample that I have provided is a *very* simple one illustrating how to instantiate the proper COM objects, and the basics of how to interface with them (in a very simple example)
标签： downloaded Assuming Setting project
上传时间： 2015-01-16
上传用户：阳光少年2016
Tug of War(A tug of war is to be arranged at the local office picnic. For the tug of war, the picnickers must be divided into two teams. Each person must be on one team or the other the number of people on the two teams must not differ by more than 1 the total weight of the people on each team should be as nearly equal as possible. The first line of input contains n the number of people at the picnic. n lines follow. The first line gives the weight of person 1 the second the weight of person 2 and so on. Each weight is an integer between 1 and 450. There are at most 100 people at the picnic. Your output will be a single line containing 2 numbers: the total weight of the people on one team, and the total weight of the people on the other team. If these numbers differ, give the lesser first. )
上传时间： 2014-01-07
上传用户：离殇
(7)--j2me软件教学，sun one studio环境
上传时间： 2014-01-12
上传用户：xlcky
WHAT MIME64 IS: MIME64 is an encoding described in RFC1341 as MIME base64.Its purpose is to encode binary files into ASCII so that they may be passedthrough e-mail gates. In this regard, MIME64 is similar to UUENCODE.Although most binaries these days are transmitted using UUENCODE, Ihave seen a few using MIME64, and I have had requests from friends thatI decode MIME64 files that have fallen into their hands. As long assome MIME64 continues to exist, a package such as this one is usefulto have.
标签： MIME described 64 encoding
上传时间： 2013-12-17
上传用户：maizezhen
说明：本程序用于矩形截面偏心受压构件对称配筋的设计和复核以及不对称配筋的复核。本程序有文件支持，但无需建立数据文件，请按屏幕提示输入数据，并注意单位。本程序可在此文件同目录上自动生成下列文件：设计数据文件“design.dat”，设计结果文件“design.out”；复核数据文件“verify.dat”，复核结果文件“verify.out”。
上传时间： 2015-03-19
上传用户：xinzhch
浮点数基本运算 浮点数的基本运算主要有四则运算、符号处理、大小比较，以及浮点数分柝等。 包含头文件 "fn.hpp" #include "fn.hpp" 浮点数基本运算 浮点数的基本运算中有加、减、乘、除、取负、绝对值、相等比较等。 加减乘除 加、减、乘、除四个运算极为相似，都是需要两个参数，结果当然也是浮点数了。 例子： // 加 减 乘 除 btil::fn::plus<f1, f2>::value // f1+f2 的结果 btil::fn::minus<f1, f2>::value // f1-f2 的结果 btil::fn::multiplies<f1, f2>::value // f1*f2 的结果 btil::fn::divides<f1, f2>::value // f1/f2 的结果 plus<f1, f2>::value::f_val // f1+f2 的结果的值 struct one { static const double f_val = 1.0 } // 两个浮点数 struct two { static const double f_val = 2.0 } minus<two, plus<divides<one, two>::value, one>::value >::value::f_val == 0.5 取负 取负运算就是取一个浮点数的负数。
上传时间： 2014-12-06
上传用户：exxxds
