小编mab*_*abg的帖子

我试图通过Firebase邀请发送邀请.从联系人中选择一封电子邮件时,Firebase表示已发送邀请,但从未收到该电子邮件.

在控制台上,SHA1认证是可配置的.

返回的错误代码始终为RESULT_OK,并且从AppInviteInvitation.getInvitationIds返回的邀请数正确无误.

SDK在gradle上使用最新版本10.0.1进行更新,如文档中所述.

创建邀请的代码是:

Intent intent = new AppInviteInvitation.IntentBuilder(title)
                .setMessage(msg)
                .setCallToActionText(callToActionText)
                .setOtherPlatformsTargetApplication(AppInviteInvitation.IntentBuilder.PlatformMode.PROJECT_PLATFORM_IOS, IOS_CLIENT_ID)
                .build();

有任何想法吗？

假设有这样一个类：

class A {
private:
   QFile file;

public:
   A::A(QFile file): file(file) {}

   void doSomething() {
      file.open(QIODevice::WriteOnly); 
      // ... do operations that can throw an exception
      file.close();
   } 
}

如果发生某些事情， close() 永远不会调用它。正确的方法是使用 try - finally，但 C++ 不支持它：

class A {
private:
   QFile file;

public:
   A::A(QFile file): file(file) {}

   void doSomething() {
      file.open(QIODevice::WriteOnly); 
      try {
          // ... do operations that can throw an exception
      }
      finally {
          file.close();
      }
   } 
}

我怎样才能在C++上做到这一点？

我正在尝试在我的iOS应用程序中使用Firebase SDK 3.7.0,但它会出现链接错误.

我已经按照README.md中描述的步骤进行了操作,它包含在从https://firebase.google.com/docs/ios/setup下载的框架SDK 上,不包含CocoaPods部分.

在链接时,出现以下错误:

Undefined symbols for architecture armv7:
  "_sqlite3_bind_double", referenced from:
      ___48-[FIRASqliteStore bindValues:toStatement:error:]_block_invoke in FirebaseAnalytics(FIRASqliteStore_2e8a0f466f3be880241569cb2066f73b.o)
  "_sqlite3_bind_null", referenced from:
      ___48-[FIRASqliteStore bindValues:toStatement:error:]_block_invoke in FirebaseAnalytics(FIRASqliteStore_2e8a0f466f3be880241569cb2066f73b.o)
  "_sqlite3_column_type", referenced from:
      -[FIRASqliteStore valueAtColumn:forStatement:] in FirebaseAnalytics(FIRASqliteStore_2e8a0f466f3be880241569cb2066f73b.o)
  "_sqlite3_column_double", referenced from:
      -[FIRASqliteStore valueAtColumn:forStatement:] in FirebaseAnalytics(FIRASqliteStore_2e8a0f466f3be880241569cb2066f73b.o)
  "_sqlite3_open_v2", referenced from:
      -[FIRASqliteStore openAndValidateDatabase:] in FirebaseAnalytics(FIRASqliteStore_2e8a0f466f3be880241569cb2066f73b.o)
  "_sqlite3_clear_bindings", referenced from:
      -[FIRASqliteStore cleanUpStatement:forQuery:] in FirebaseAnalytics(FIRASqliteStore_2e8a0f466f3be880241569cb2066f73b.o)
      -[FIRASqliteStore bindValues:toStatement:error:] in FirebaseAnalytics(FIRASqliteStore_2e8a0f466f3be880241569cb2066f73b.o)
  "_sqlite3_column_count", referenced from:
      -[FIRASqliteStore recordsForQuery:parameterValues:error:withFilter:] in FirebaseAnalytics(FIRASqliteStore_2e8a0f466f3be880241569cb2066f73b.o)
      -[FIRASqliteStore validateDatabaseWithError:] in FirebaseAnalytics(FIRASqliteStore_2e8a0f466f3be880241569cb2066f73b.o)
  "_OBJC_CLASS_$_SSReadingList", referenced from:
      objc-class-ref in GoogleSignIn(please_link_SafariServices.framework_.o)
  "_OBJC_CLASS_$_MFMessageComposeViewController", referenced from:
      objc-class-ref …

我有一个名为Symbol的类,我想创建一个QVector子类(Symbols)来添加一些有用的方法.但是当我使用来自另一个类A的符号时,编译器会给出错误'符号没有命名类型'.

class Symbols: public QVector< Symbol >
{
public:
    Symbols() {}

    // Useful methods
    QSymbol findSymbol(const QString &name);

    // ...
};

class A
{
private:
    Symbols symbols;
};

它是否正确分类？

为什么在编译A类时出现'符号不命名类型'？