大约有 43,000 项符合查询结果（耗时：0.0187秒） [XML]
Should Gemfile.lock be included in .gitignore?
...ndler docs address this question as well:
ORIGINAL: http://gembundler.com/v1.3/rationale.html
EDIT: http://web.archive.org/web/20160309170442/http://bundler.io/v1.3/rationale.html
See the section called "Checking Your Code into Version Control":
After developing your application for a while, ...
深入理解 x86/x64 的中断体系 - 操作系统(内核) - 清泛网 - 专注C/C++及内核技术
...间低端，实际上等于 ROM 被映射到地址空间最高端和低端位置。
此时在系统的内存里其实并不存在 BIOS 代码，ROM BIOS 的一部分职责是负责安装 BIOS 代码进入系统内存。
jmp far f000:e05b
典型是这条指令就是 0xFFFFFFF0 处...
awk without printing newline
...
awk '{sum+=$3}; END {printf "%f",sum/NR}' ${file}_${f}_v1.xls >> to-plot-p.xls
print will insert a newline by default. You dont want that to happen, hence use printf instead.
share
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创业公司倒闭大潮 教你正确烧钱速度？ - 资讯 - 清泛网 - 专注C/C++及内核技术
...标是投资工程师团队保持产品领先、投资新办公室/地理位置(提前抢占市场)、投资市场(抢先吸引顾客)，而所有这些事经常需要在你赚钱前就大量花钱。不过如果你使用这一策略，很大程度将必须依赖下面我要讲的第3点。
资金...
How to wait for several Futures?
...lt;- func2.concurrently
f3 <- func3.concurrently
} yield for {
v1 <- f1
v2 <- f2
v3 <- f3
} yield (v1,v2,v3)
}.future
f.onFailure { case t => println("future failed $t") }
In the example above, f1,f2 and f3 will run concurrently and if any fail in any order the futu...
What is the best way to detect a mobile device?
...
100
Why not: if( screen.width <= 480 ) { // is mobile }
– andrewrjones
Jan 20 '13 at 1:05
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linux 下巧妙使用squid代理服务器 - 更多技术 - 清泛网 - 专注C/C++及内核技术
...下载加速服务：
四、Web缓存的类型和特点:
Web缓存的位置可以有三种，一是可以防止在客户端，二是放在服务器，三是放在客户端与服务器之间的某个网络节点上，这个网络节点就是web代理服务器。
1 客户端缓存
2 代理服务...
How to clone all remote branches in Git?
...-a
* master
remotes/origin/HEAD
remotes/origin/master
remotes/origin/v1.0-stable
remotes/origin/experimental
If you just want to take a quick peek at an upstream branch, you can check it out directly:
$ git checkout origin/experimental
But if you want to work on that branch, you'll need to...
Index all *except* one item in python
...at if variable is list of lists, some approaches would fail.
For example:
v1 = [[range(3)] for x in range(4)]
v2 = v1[:3]+v1[4:] # this fails
v2
For the general case, use
removed_index = 1
v1 = [[range(3)] for x in range(4)]
v2 = [x for i,x in enumerate(v1) if x!=removed_index]
v2
...
Good Java graph algorithm library? [closed]
...new SimpleGraph<String, DefaultEdge>(DefaultEdge.class);
String v1 = "v1";
String v2 = "v2";
String v3 = "v3";
String v4 = "v4";
// add the vertices
g.addVertex(v1);
g.addVertex(v2);
g.addVertex(v3);
g.addVertex(v4);
// add edges to create a circuit
...