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Intel SMID指令集编译错误: inlining failed in call to always_inline \'x...

Intel SMID指令集编译错误: inlining failed in call to always_inline 'xxx': target specific option mismatch xxxintel_smid_compile_error最近在使用CPU指令集优化代码的时候,编译出错,报错如下: usr lib64 gcc x86_64-suse-linux 7 include avx2intrin h:252:1: error: inlining ...
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socks5代理工作流程及技术原理 - 更多技术 - 清泛网 - 专注C/C++及内核技术

socks5代理工作流程及技术原理socks5-proxy-protocolsocks5协议是一款广泛使用的代理协议,它在使用TCP IP协议通讯的前端机器和服务器机器之间扮演一个中介角色,使得内部网中的前端机器变得能够访问Internet网中的服 一、socks5协议 ...
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eclipse升级后启动失败:Heap堆内存不足 - 更多技术 - 清泛网 - 专注C/C++及内核技术

...eap堆内存不足eclipse_upgrade_outof_memory修改 eclipse ini,更改 -Xmx521m 为 -Xmx3g 本来打算从Eclipse(2020-06)更新一下(2021-03),没想到升级成后Eclipse启动不了,错误日志如下: !SESSION 2021-06-16 17:03:13.186 -----------------------------------------...
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PHP 安装 ZIP 扩展 - 更多技术 - 清泛网 - 专注C/C++及内核技术

PHP 安装 ZIP 扩展php_zip_extension一、安装libzipcd usr local srcwget https: libzip org download libzip-1 3 2 tar gztar zxf libzip-1 3 2 tar gzcd libzip-1 3 2 configuremake && make install如 一、安装libzip cd /usr/local/src wget https://libzip.org/download/libzip-1.3.2.tar.gz ...
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背包:将代码块复制并粘贴到不同的屏幕和项目 · App Inventor 2 中文网

...初始化将自动重命名。 例如,如果你尝试粘贴全局变量 X 的变量初始化块,并且工作区已包含 X 的初始化块,则该变量将在粘贴的块中重命名为 X2。 同样,如果你尝试粘贴过程 foo 的过程定义,并且工作区已包含 foo 的定义,...
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BLE蓝牙 定时功能 - App Inventor 2 中文网 - 清泛IT社区,为创新赋能!

...启用计时设置为 真,这时才开始计时!假设定一小时,60X360X1000 这样填代码块么?引用: 逍遥虎 发表于 2024-03-23 07:48 假设定一小时,60X360X1000   这样填代码块么? 你这是 6小时。一小时以主贴截图代码块为准!谢谢指...
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模拟器启动失败:找不到 msvcp140.dll - App Inventor 2 中文网 - 清泛IT社...

msvcp140.dll 及相关的微软的运行环境安装即可: https://aka.ms/vs/16/release/vc_redist.x64.exe https://aka.ms/vs/16/release/vc_redist.x86.exe
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SQL selecting rows by most recent date

... can use a GROUP BY to group items by type and id. Then you can use the MAX() Aggregate function to get the most recent service month. The below returns a result set with ChargeId, ChargeType, and MostRecentServiceMonth SELECT CHARGEID, CHARGETYPE, MAX(SERVICEMONTH) AS "MostRecentServiceMon...
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How to print a linebreak in a python function?

...= ['a1', 'a2', 'a3'] >>> B = ['b1', 'b2', 'b3'] >>> for x in A: for i in B: print ">" + x + "\n" + i Outputs: >a1 b1 >a1 b2 >a1 b3 >a2 b1 >a2 b2 >a2 b3 >a3 b1 >a3 b2 >a3 b3 Notice that you are using /n which is not correct! ...
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String is immutable. What exactly is the meaning? [duplicate]

...are immutable? Well to your surprise, it is. When the above statement is executed, the VM takes the value of String str, i.e. "knowledge" and appends " base", giving us the value "knowledge base". Now, since Strings are immutable, the VM can't assign this value to str, so it creates a new String ob...