登录
首页 » Java » ojdbc14-10.2.0.4.0

ojdbc14-10.2.0.4.0

于 2020-06-22 发布 文件大小:1440KB
0 60
下载积分: 1 下载次数: 0

代码说明:

  oracle ojdbc drivers 14-10

下载说明:请别用迅雷下载,失败请重下,重下不扣分!

发表评论

0 个回复

  • auz-procedures-procedures
    一个不错的DES加密和解密程序,可以看看这个程序(A good DES encryption and decryption procedures, can take a look at this program)
    2017-08-21 06:27:26下载
    积分:1
  • 该游戏是使用VC++平台开发的简单游戏,可供游戏爱好者学习
    该游戏是使用VC++平台开发的简单游戏,可供游戏爱好者学习
    2023-02-11 23:15:03下载
    积分:1
  • 9nine
    复制 测量长度 比较 等函数的 书写 还有 一些其他函数的书写 都是c++课程老师布置的作业(Copy measuring the length of comparison function of writing, there are some other functions of writing are c++ course teacher assignments)
    2012-08-06 22:52:46下载
    积分:1
  • 用锁文件实现进程通信
    用锁文件实现进程通信-lock document to achieve communication proce
    2022-05-21 20:36:29下载
    积分:1
  • distribution-line-lightning
    电力系统配电线路防雷设计指导书,用于雷击风险评估,防雷措施效果计算。(Power system distribution line lightning protection design guide books, the risk assessment for lightning, lightning effects of measures calculated.)
    2013-11-10 12:15:22下载
    积分:1
  • Empirical mode decomposition process (matlab)
    经验模式分解程序(matlab)可以对输入信号进行经验模式分解-Empirical mode decomposition process (matlab)
    2022-07-16 23:40:09下载
    积分:1
  • 一篇对于催眠术的说明及基本教程,详细阐述了相关概念和学习步骤!...
    一篇对于催眠术的说明及其基本教程,详细阐述了相关概念和学习步骤!-For a description of hypnosis and its basic curriculum, elaborated on related concepts and learning the steps!
    2022-01-28 01:13:35下载
    积分:1
  • 东南大学C++课件,讲得真的很不错,很适合初学者
    东南大学C++课件,讲得真的很不错,很适合初学者-Southeast University, C++ software, put it quite true, it is suitable for beginners
    2023-03-06 22:35:03下载
    积分:1
  • ILD-nodeMCU开发板原理图
    onenet 开发板原理图 物联网开发 esp8266(Onenet development board schematics, Internet of things development esp8266)
    2018-06-20 08:26:31下载
    积分:1
  • cop2000
    说明:  cop2000实现无符号数乘法——两个8位无符号数相乘,所乘结果是16位,采用原码一位乘,在计算时,用乘数寄存器的最低位来控制部分积是否与被乘数相加,然后右移部分积和乘数,同时乘数寄存器接收部分积右移出来的一位,完成运算后,部分积寄存器保存乘积的高位部分,乘数寄存器中保存乘积的低位部分。(Cop2000 implements unsigned number multiplication-multiplication of two 8-bit unsigned numbers. The multiplied result is 16 bits. When calculating, the lowest bit of multiplier register is used to control whether or not the partial product is added to the multiplier, and then the partial product and multiplier are moved right. At the same time, the multiplier register receives the right-shifted bit of the partial product. After completing the operation, the partial product register saves the product. The higher part, the lower part of the product is saved in the multiplier register.)
    2021-01-03 19:28:55下载
    积分:1
  • 696524资源总数
  • 103945会员总数
  • 46今日下载