iOS 9 Command failed due to signal: Segmentation fault: 11

Question

我最近升级到Xcode 7。我的代码在6.3和iOS 8.4上运行良好，现在将无法编译分段错误。

我正在使用SQLite的Stephencelis库。似乎是它引起了问题。

我将感谢任何帮助，任何想法。下面是错误消息的最后一部分:

1. 在/Users/luben/Downloads/SQLite.swift-master/SQLite/Query.swift:740:22
2. 当类型检查声明0x7fbab3ee2d68在/Users/luben/Downloads/SQLite.swift-master/SQLite/Query.swift:740:51

3. While类型检查表达式在[/Users/luben/Downloads/SQLite.swift-master/SQLite/Query.swift:740:51 - line:774:7] RangeText="{var (columnNames, idx) = (String: Int, 0)列:每个在self.query.columns ??[Expression(literal: "*")] {let pair = each.expression.SQL.characters.split {$0 == "."}。map {String($0)}let (tableName, column) = (pair。计数> 1 ?对。第一:nil, pair.last!

       func expandGlob(namespace: Bool) -> Query -> Void {
           return { table in
               var query = Query(table.database, table.tableName.unaliased)
               if let columns = table.columns { query.columns = columns  }
               var names = query.selectStatement.columnNames.map { quote(identifier: $0) }
               if namespace { names = names.map { "(table.tableName.SQL).($0)" } }
               for name in names { columnNames[name] = idx++ }
           }
       }
       if column == "*" {
           let tables = [self.query.select(*)] + self.query.joins.map { $0.table }
           if let tableName = tableName {
               for table in tables {
                   if table.tableName.SQL == tableName {
                       expandGlob(true)(table)
                       continue column
                   }
               }
               assertionFailure("no such table: (tableName)")
           }
           tables.map(expandGlob(self.query.joins.count > 0))
           continue
       }
       columnNames[each.expression.SQL] = idx++
   }
   return columnNames
   

   } ()"

Answer 1

GitHub上的SQLite.swift项目提供了一个Swift 2分支。这也许能解决问题。