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trail
- 是否保留重测结果,可以根据自己实际需求而做选择性的保存测试结果-save the retest data result
ABC
- 基于asm编译器的简单汇编程序，功能：用8086的16位机完成两个32数的相乘，随意给两个32位数，输出相乘的结果-Asm compiler based on a simple assembler, features: 16-bit machine with a 8086 completion of two numbers multiplied by 32, 32 at random to two digits, the result of multiplying the output
pks-source
- Web broker demo. According to financial result of company, program calculate racio, scoring, rrl, roa, ebit, ebitda and other financial parameters.
semantic_analyse_program
- 用C++编写的语义分析程序，基于LR文法，能实现词法分析，语法分析，返回相应结果，并返回归约过程和四元式。并附有实验报告和源工程。-Written in C++ semantic analysis program, based on the LR grammar, to achieve lexical analysis, parsing, returns the appropriate result, and return to the process of reduction and quat
busy
- 利用ICL7135的BUSY怎样读取A/D转换结果-How to read using the BUSY ICL7135 A/D conversion result
Speakerrecognitionbased-on-MFCC
- 这是一篇工学硕士论文，主要说关于说话人语音识别，，以支持向量机为分类器，分别以MFCC和IM-FCC为特征单独执行分类，将得到结果按某种方式融合，取两者之长，最后做出判决来提高说话人识别系统的性能。-Speaker recognition is a kind of technology that take the use of the information contained in the speech signal to identify who is the speaker or to
match-words
- 实现两个字符串之间的比较，并将最终结果打印出来。-Achieve a comparison between two strings and print the final result.
Digital-matrix-program
- 该程序实现显示一个数字的方阵，程序中的清屏功能可将上次运行的结果 清除，重新显示下次运行结果。本程序在dos中或windows98中运行。-The program displays a number of the square to achieve the program s clear-screen function can be the result of the last run Clear, re-display the results next time. The progr
random
- 本程序需在DOS下运行。每运行一次可执行程序random.exe，即可显示任意两个 16进制数的加法或减法表达式及其运算结果。在减法运算中，如果被减数小于减数，显示 “Divide Overflow”的提示信息。-The process required to run under DOS. Each time you run the executable program random.exe, you can display any two hexadecimal expression
ptcd
- 这是一个计算数字平台长度的程序，如： 运行时输入：111122333，结果为:4 1111 注：本程序在dos中或windows98中运行-This is a procedure for calculating the length of digital platforms, such as: run-time type: 111 122 333, the result is: 41111 Note: This program is run dos or windows98
Communication-Assignment
- simulation result of the am receiver circuit used to demodulate the modulated signal, it contain the report and out put result
masm5
- 在数据段DATA中有两个数据X和Y，假设X=1122H，Y=3344H，编程求两个字的和，结果存放到Z单元中。-DATA in the data segment has two data X and Y, assuming X = 1122H, Y = 3344H, program requirements and the word, the result in the Z unit.
sor
- SOR ap for CUDA, iterate many times and proceed reduction for check the result of program.
BCDapplication
- 1、 把非压缩的BCD码3532H转变为压缩的BCD码52H，并输出到屏幕。 2、使用BCD码校正指令，实现两个4位十进制数的加法4678+2556，并把结果输出到屏幕。 -1, the non-compressed into a BCD code 3532H compressed BCD code 52H, and output to the screen. 2, the BCD code of correction commands to achieve two four decim
2006_book-TransactionalMemory
- 电子书，综述了事务存储，事务存储是一种很有前途的并行编程模型-ABSTRACT The advent of multicore processors has renewed interest in the idea of incorporating transactions into the programming model used to write parallel programs. This approach, known as transactional memory,
ADD_BCD
- This routine performs a 2 Digit Unsigned BCD Addition It is assumed that the two BCD numbers to be added are in locations Num_1 & Num_2. The result is the sum of Num_1+Num_2 and is stored in location Num_2 and the overflow carry is returned
BCD_ADD
- This routine performs a 2 Digit Unsigned BCD Addition It is assumed that the two BCD numbers to be added are in locations Num_1 & Num_2. The result is the sum of Num_1+Num_2 and is stored in location Num_2 and the overflow carry is returned
BCD_SUB
- *** *** *** * Unsigned BCD Subtraction *** *** *** This routine performs a 2 Digit Unsigned BCD Subtraction. It is assumed that the two BCD numbers to be subtracted are in locations Num_1 & Num_2. The result is the difference of Num_1 & Num_
paixu
- 自己写的8086实现数组的排序算法，采用冒泡法，并可将结果输出-Written in 8086 to achieve their own array of sorting algorithms, using the bubble method, and output the result
leijia
- （2） 用汇编程序实现1～100的加法运算 1. 设置循环变量和存储运算结果的寄存器的初值 2. 利用循环结构实现累加 3. 验证结果（0x13BA）。 -) compilation of procedures used to achieve 1 to 100 addition operationsThe 1 set of the loop variable and storage operation result register value2 using the loop s