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LCM_GPM1002B0_2
GPM1002B0 初始化代码，颜色艳丽，可以直接使用(GPM1002B0 initialization code)
- 2014-01-15 10:46:35下载
- 积分：1
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femsrc
用C/C++编程求解了一个有限元算例二维传热问题,可以作为有限元学习的编程实例，以更好地理解有限元理论，并为进一步使用大型有限元软件打下基础。(With C/C++ Programming to solve a finite element numerical example of two-dimensional heat transfer problems, finite element can be used as examples to learn programming in order to better understand the finite element theory, and for the further use of large-scale finite element software to lay basis.)
- 2008-06-28 12:51:20下载
- 积分：1
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TEMEX
MATLAB (MATLAB )
- 2015-01-01 17:10:42下载
- 积分：1
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MATLABexperiment
基于matlab平台的编程 实现pcm 编码的功能(the file is based on the matlab and it has the function of pcmcoding )
- 2009-05-29 09:18:08下载
- 积分：1
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AfxGetApp
AfxGetApp()应用，比较简单的说明，我从网上找的。(AfxGetApp (), relatively simple explanation, I find from the Internet.)
- 2006-12-28 08:38:58下载
- 积分：1
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DSP_FFT
产生两不迥频率的信号，并对两信号作复数FFT变换，找出频率为50Hz的信号对其进行能量计算(Two distant signal is not generated frequency, and two signals for complex FFT transform to identify the frequency of 50Hz signal to calculate its energy)
- 2013-12-01 10:35:35下载
- 积分：1
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boqingwen
对二阶rc等效电池模型运用matlab进行最小二乘法在线系统辨识的程序(Equivalent second-order rc battery model using least squares method matlab online system identification procedures
)
- 2020-09-12 16:28:01下载
- 积分：1
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Boost-develop-guide
Boost程序库完全开发指南,学习boost的必备，初学者最好看一看(Boost library fully developed guidelines, learning the necessary boost, take a look at the beginner)
- 2014-11-15 09:58:04下载
- 积分：1
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EmailParser
Email parser in C# dotnet
- 2013-12-28 23:59:00下载
- 积分：1
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5
一道程序编译顺序的考题，涉及到函数调用的先后顺序及运算符号的优先级等问题。下面我展开给你讲。
C的程序编译总是从main函数开始的，这道题的重点在“fun（（int）fun（a+c,b）,a-c）） ”语句。
系统首先要确定最外层 fun()函数的实参，第一个参数的确定需要递归调用fun()函数（不妨称其为内层函数）。内层函数的两个参数分别为x=a+b=2+8=10、y=b=5,执行函数体x+y=10+5=15,于是得外层函数的参数x=15。其另一个参数y=a-c=2-b=-6,再次执行函数体,得最终返回值x+y=15+(-6)=9。 (Compiling together the sequence of test procedures, involving the sequence of function calls and operator symbols, such as the priority problem. Now I give you to start speaking. Procedures for C compiler always start from the main function and at这道题the focus of " fun ((int) fun (a+ c, b), ac)) " statement. System must first determine the most outer layer of fun () function of real parameters, the first parameters of recursive calls required fun () function (may be called the inner function). Inner function separately for the two parameters x = a+ b = 2+8 = 10, y = b = 5, to execute the function body x+ y = 10+5 = 15, then the outer function parameters were x = 15 . Its another parameter y = ac = 2-b =- 6, once again to execute the function body may eventually return the value of x+ y = 15+ (-6) = 9.)
- 2009-03-15 15:36:23下载
- 积分：1
