Question

我目前正在开发一个与sqlite数据库交互的项目。问题是每次我想连接数据库时，我都要打开并准备数据库。所以我想让这些步骤更加通用，我决定创建一个通用类，为我做这些步骤。

+(void)openAndPrepareDatabase:(sqlite3 *)db andStatement:(sqlite3_stmt *)statement andSql:(const char *)sql
{
@try
{
    if(!sqlite3_open([[self getDatabasePath] UTF8String], &db) == SQLITE_OK)
    {
        NSException *myException = [NSException exceptionWithName:@"Can't open database" reason:@"Can't open database" userInfo:Nil];

        @throw myException;
    }

    if(!sqlite3_prepare(db, sql, -1, &statement, NULL) == SQLITE_OK)
    {
        NSException *myException = [NSException exceptionWithName:@"Can't prepare database" reason:@"Can't prepare database" userInfo:Nil];

        @throw myException;
    }
}
@catch (NSException *exception)
{
    @throw exception;
}
}


+(void)openAndPrepareDatabaseV2:(sqlite3 *)db andStatement:(sqlite3_stmt *)statement andSql:(const char *)sql
{
@try
{
    if(!sqlite3_open([[self getDatabasePath] UTF8String], &db) == SQLITE_OK)
    {
        NSException *myException = [NSException exceptionWithName:@"Can't open database" reason:@"Can't open database" userInfo:Nil];

        @throw myException;
    }

    if(!sqlite3_prepare_v2(db, sql, -1, &statement, NULL) == SQLITE_OK)
    {
        NSException *myException = [NSException exceptionWithName:@"Can't prepare database" reason:@"Can't prepare database" userInfo:Nil];

        @throw myException;
    }
}
@catch (NSException *exception)
{
    @throw exception;
}
}

但是当我试图在我的对象中调用它时，I.E：

[Common openAndPrepareDatabase:&db andStatement:&statement andSql:sql];

我收到了警告：

"Incompatible pointer types sending 'sqlite3_stmt **' (aka 'struct sqlite3_stmt **')  to parameter of type 'sqlite3_stmt *' (aka 'struct sqlite3_stmt *'); remove &"

有谁知道我的问题的解决方案？

Answer 1

有谁知道我的问题的解决方案？

编译器 只显示了一个。

“不兼容的指针......等等; 删除＆amp; ”

（强调我的）

在目标C中传递参考参数

1 个答案: