0

当我尝试构建和分析事物时,下面的代码片段会泄漏。

这段代码有什么问题,请告诉我

- ( NSString *) getSubCategoryTitle:(NSString*)dbPath:(NSString*)ID{
        NSString *subCategoryTitle;

    if (sqlite3_open([dbPath UTF8String], &database) == SQLITE_OK) 
    {
        NSString *selectSQL = [NSString stringWithFormat: @"select sub_category_name from sub_categories where id = %@",ID];

        NSLog(@"%@ I am creashes here", selectSQL);

        const char *sql_query_stmt = [selectSQL UTF8String];

        sqlite3_stmt *selectstmt;

        if(sqlite3_prepare_v2(database, sql_query_stmt, -1, &selectstmt, NULL) == SQLITE_OK) 
        {

            while(sqlite3_step(selectstmt) == SQLITE_ROW) 
            {

                subCategoryTitle = [[NSString alloc] initWithUTF8String:
                                    (const char *) sqlite3_column_text(selectstmt, 0)];


            }
        }

        sqlite3_finalize(selectstmt);    

    }   
    sqlite3_close(database); 


    return [subCategoryTitle autorelease];
}
4

1 回答 1

2

您将实例分配到subCategoryTitle循环中,但不要释放先前的分配。

subCategoryTitle = [[NSString alloc] initWithUTF8String:
                                (const char *) sqlite3_column_text(selectstmt, 0)];

要么(自动)释放它,要么直接转到最后一行,并避免这种情况,因为它没有多大意义。

仅创建最后一个对象的示例:

char * col_text = NULL;
while(sqlite3_step(selectstmt) == SQLITE_ROW) 
{
    col_text = sqlite3_column_text(selectstmt, 0);
}
if (col_text != NULL)
{
    subCategoryTitle = [[NSString alloc] initWithUTF8String:col_text];
}
于 2012-05-11T06:07:32.017 回答