我想只获得预处理版本file.cc.我做了g++ -E file.cc,得到了:
# 1 "file.cc"
# 1 "<command-line>"
# 1 "file.cc"
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我做错了什么?
我正在开发应用程序并共享扩展并尝试使用核心数据.但是当我在扩展中插入项目时,这些项目仅在扩展中可见但不是从容器应用程序中可见(例如,我从应用程序执行NSFetchRequest并获得零项目,但在app中我得到> 0).我使用以下代码获取持久性容器:
lazy var persistentContainer: NSPersistentContainer = {
let container = NSPersistentContainer(name: "appname")
container.loadPersistentStores(completionHandler: { (storeDescription, error) in
if let error = error {
fatalError("Unresolved error \(error)")
}
})
return container
}()
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此外,还会检查appname.xcdatamodeld的目标成员应用程序和扩展程序.如何正确共享容器应用和扩展的核心数据?
我正在尝试使用ubuntu在ec2实例上安装gcc
sudo apt-get install gcc
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给我:
Do you want to continue? [Y/n]
Err http://us-west-2.ec2.archive.ubuntu.com/ubuntu/ trusty-updates/main linux-libc-dev amd64 3.13.0-36.63
404 Not Found [IP: 91.189.88.149 80]
Err http://security.ubuntu.com/ubuntu/ trusty-security/main linux-libc-dev amd64 3.13.0-36.63
404 Not Found [IP: 91.189.92.201 80]
E: Failed to fetch http://security.ubuntu.com/ubuntu/pool/main/l/linux/linux-libc-dev_3.13.0-36.63_amd64.deb 404 Not Found [IP: 91.189.92.201 80]
E: Unable to fetch some archives, maybe run apt-get update or try with --fix-missing?
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我错过了什么?
我想将指向成员的指针作为foo1的模板参数.这是代码:
struct baz{
int qux;
};
template<typename C, typename T, T C::*m>
struct foo1{};
template<typename C, typename T>
void barr2(T C::*m){
}
template<typename C, typename T>
void barr1(T C::*m){
barr2(m); // ok
foo1<C, T, &baz::qux> _; // ok
foo1<C, T, m> f; // g++4.6.1 error here; how to pass 'm' correctly ?
}
int main(){
barr1(&baz::qux);
}
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那应该怎么样?
问题是我无法在接收器中获得推荐人uri.在清单文件中,我有类似的东西:
<receiver
android:name="app.InstallReceiver"
android:exported="true" >
<intent-filter>
<action android:name="com.android.vending.INSTALL_REFERRER" />
</intent-filter>
</receiver>
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Adb广播命令:
adb shell am broadcast -a com.android.vending.INSTALL_REFERRER
-n app/app.InstallReceiver --es "referrer" "utm_source=testSource&utm_medium=testMedium&utm_term=testTerm&utm_content=testContent&utm_campaign=testCampaign"
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在InstallReceiver类中:
@Override
public void onReceive(Context context, Intent intent) {
Uri uri = intent.getData(); // getting uri is null
String referrer = intent.getStringExtra("referrer");
// referrer is only contains "utm_source=testSource" and no more
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所以问题是完整的推荐人和我的意图有什么不对.
在此片段中,delete返回true.那么为什么在删除后成功调用f?
function X() {
this.f = function() {
console.log("X::f");
}
}
x = new X;
var f = x.f.bind(x);
console.log("delete: " + delete x);
f();Run Code Online (Sandbox Code Playgroud)
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