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CMakeLists指定链接顺序(error adding symbols: DSO missing from command line)

转载 https://www.cnblogs.com/Hocker/p/8269506.html https://www.cnblogs.com/OCaml/archive/2012/06/18/2554086.html#sec-1-1   关于链接库的顺序问题,我看了下链接库的顺序问题一文,正常的情况下,链动态接库时是从左至右开始链接,如果出现了顺序错误,gcc自动帮我们重定位,我认为保

linux zip 命令

zip命令: zip命令是压缩和文件打包的工具,unzip是解压的命令。 zip命令将一个或多个文件压缩成一个zip文件(文件大于4G时zip64默认后缀),整个目录也可以被压缩到一个文件中。text文件的压缩比例是2:1或3:1。 ZIPOPT环境变量 执行zip时会调用ZIPOPT的环境变量,如果我们想定制一些操作

git add失败

git 报错:error: insufficient permission for adding an object to repository database .git/objects 进入到你项目中的.git目录中(切换到root) chgrp -R groupname . chmod -R g+rwX .最后再切换回你的用户

D. Not Adding

D. Not Adding 思路: 我们可以枚举每一个\(1-10^6\)每一个整数,判断它们是否合法,若当前数在数组里面且原数组里面没有任意两个它的倍数的\(gcd\)等于它为不合法的情况。 时间复杂度:\(O(n + maxn\ln(maxn))\) 代码: #include <bits/stdc++.h> using namespace std; int n; vector<in

Step 3: Adding Usage Requirements for a Library

1添加库的使用规范 2具体脚本 2.1target_compile_definitions() 向工程中加入预处理定义 语法: target_compile_definitions(<target> <INTERFACE|PUBLIC|PRIVATE> [items1...] [<INTERFACE|PUBLIC|PRIVATE> [items2...] ...]) cmake --build . --config release : 可以生

Step 2: Adding a Library

参考 参考 1完整代码 |-step2 |--CMakeLists.txt |--tutorial.cxx |--MathFunctions |--CMakeLists.txt |--mysqrt.h |--mysqrt.cxx |--CMakeLists.txt cmake_minimum_required(VERSION 3.10) project(Tutorial) # 设置宏变量 option(U

[BUUCTF-pwn] bssidessf_ctf_2020_adding_machine

难度不大,要求输入n个数,但栈空间只能放127个数,要求输入数<127但可以输入负数,其它地址在输入debug时可以全部得到。 这里由于没有中途退出机制,需要精确的输入一个负数表示执行次数 v6 = get_long("Number of numbers to add"); if ( v6 <= 127 ) { get_data(v8, (unsign

Cannot create Launcher without at least one TestEngine; consider adding an engine implementation JAR

当在IDEA 使用gradle创建工程时,默认是使用junit5的,但现在我想切换为junit4,于是便把依赖里junit5的包全部删除,然后导入junit4的包,这时若是执行测试代码就会出现标题上的错误,该如何解决呢,还要删除一处地方,即如下图位置代码注释即可  

command zip/unzip

zip/unzip 用来压缩和解压缩文件。 压缩 a.log b.log $touch {a..e}.log $ls a.log b.log c.log d.log e.log $zip test.zip a.log b.log adding: a.log (stored 0%) adding: b.log (stored 0%) $ls a.log b.log c.log d.log e.log test.zip 在test.zip里追加 c.l

Adding a new SSH key to your GitHub account

  完美文档: https://docs.github.com/en/github/authenticating-to-github/connecting-to-github-with-ssh/adding-a-new-ssh-key-to-your-github-account step1: 在机器上生成ssh key; step2: 添加到github.

萌新web2

  This is similar to 萌新web1, but the only difference between them is here adding the function preg_match() that is a regular expression to filter 'or' and '+'.  So we just construct the payload without the above stuff.   

Consider enabling transient error resiliency by adding 'EnableRetryOnFailure()' to the �

Consider enabling transient error resiliency by adding 'EnableRetryOnFailure()' to the 'UseMySql' call ef core连接不上MySQL 连接字符串是: 'Data Source = 127.0.0.1;Database = test1;UserID=root;password=123456;pooling=true;port=3306;sslmode=n

Adding Applications for JavaScript Storefronts in SAP 电商云

帮助文档: https://help.sap.com/viewer/b2f400d4c0414461a4bb7e115dccd779/v2011/en-US/63577f67a67347bf9f4765a5385ead33.html 从 Github sample Repository 拷贝出来,js-storefront 文件夹下面有个 Spartacusstore 文件夹,进入其 manifest.json 文件,可以在 applications 数组里添

Idea 热部署

  1. Adding devtoolls to your project <dependency> <groupId>org.springframework.boot</groupId> <artifactId>spring-boot-devtools</artifactId> <scope>runtime</scope> <optional>true</optional>

如何安装Python3.8版本的TensorFlow?

现在tensorflow 2.4 支持 3.6 3.7 3.8 可以放心安装 pip install tensorflow  Downloading https://mirrors.aliyun.com/pypi/packages/59/9b/tensorflow-2.5.0-cp38-cp38-manylinux2010_x86_64.whl (454.4 MB) 升级的话可以加个 --upgrade pip install tensorflow --upgrad

Spire.Office for Java 4.4.6 Crack 笑开颜

Version: 4.4.6 We are happy to announce the release of Spire.Office for Java 4.4.6. This version brings some new features, for example, Spire.XLS for Java supports adding and removing digital signature and adds a new method to customize paper size; Spire.

论文笔记之Removing the Background by Adding the Background: Towards Background Robust Self-supervised Vid

提出背景擦除(Background Erasing)方法来减轻模型对背景的依赖,从而使模型更关注动作变化。 CVPR2021 论文地址:https://arxiv.org/abs/2009.05769 1. 总述 自监督学习通过对数据本身的监督,在提高深层神经网络的视频表现能力方面显示出巨大的潜力。然而,目前的一些方法往往会存

Codechef March Challenge 2021 Div2 Consecutive Adding(CONSADD)

Codechef March Challenge 2021 Div2 Consecutive Adding(CONSADD) 题目大意: 给定两个\(n\times m\)矩阵\(A\),\(B\)和一个常数\(x\) 现在对于\(A\)操作,每次可以选择一行或者一列连续的\(x\)个,一起改变同一个数值\(v\in \Z\) 判断是否可以由\(A\)变成\(B\) 显然可以先将\(A,B\)作

Adding Images

refer to: https://www.udemy.com/course/the-web-developer-bootcamp image source: https://source.unsplash.com/ models/campground.jsconst mongoose = require('mongoose'); const Schema = mongoose.Schema; const CampgroundSchema = new Schema({ ti

1_adding basic styles

refer to: https://www.udemy.com/course/the-web-developer-bootcamp (a new ejs tool for layout: https://github.com/JacksonTian/ejs-mate) npm i ejs-mate const ejsMate = require('ejs-mate'); app.engine('ejs', ejsMate); touch views/lay

Adding a Progress Bar in R

1. Introduction I have a R program reads and revises data with dplyr::mutate(). Normally this will not be a problem. But my data frame is very large and the process logic is somehow complicated, so each time it runs will take 1-2 minutes.From time to time

AGC033F Adding Edges

一棵树\(T\)和一张图\(G\),现在对图进行加边操作:每次找到\((a,b,c)\)满足\((a,b),(b,c)\in E_G\),且\(a,b,c\)任意顺序在\(T\)上排列在一条链上。 问对图\(G\)操作到不能操作时,\(|E_G|\)是多少。 \(n,m\le 2000\) 神仙题。。。对着三个标切的,下次遇到估计还是不会做。。。 如果\(a,

October Challenge 2020 Division 1

打了两次div2,终于能打div1了,还是挺兴奋的 Positive AND(10.2) Replace for X(10.3) Inversions(10.4) D-Dimensional MST(10.4) Adding Squares(10.4) Compress all Subsegments(10.5) Rooted Minimum Spanning Tree(10.5) Random Knapsack(10.8) Village Road Network(未做) 题解赛后

git 服务器同步代码错误 error: insufficient permission for adding an object to repository database .git/object

* branch master -> FETCH_HEAD error: insufficient permission for adding an object to repository database .git/objects fatal: failed to write object fatal: unpack-objects failed git目录无写权限 我的解决方法,直接在xshell上面cd到自己的项目代码,然

CF1312C Adding Powers

题目链接:https://codeforces.com/contest/1312   题目大意: 能否对一个数组执行任意次操作,使得其变为目标数组。 对于第i次操作,我们可以放弃,或给数组中任意一个元素加上k^i   想法: 我们不难发现一个数 = k^x + k^y + k^z + ... (x != y != z) 这个形式很像我们的二进制所以我们可