我有一个关于如何分配类成员(setter)的新手问题。我习惯于编写脚本,而且大部分是通过(在python中)完成的
def set_mymember(mymember):
self.mymeber = mymember
我的同事告诉我“自我”和C ++中不需要“这个”,“这个”存在并且在这种背景下没有错,但这对我来说很难理解,所以他说我不应该关心。所以我首先按照他的建议尝试:
我的类定义: - (它应该创建一个sql查询字符串)
class Query
{
public:
Query() { }
~Query() { }
void setoptions( std::string qtext_where="", bool qtext_erl=true, std::vector<std::string> kids=std::vector<std::string>() );
Query build_query( );
void set_db_table( std::string db_table );
void set_db_key( std::string db_key );
void set_m_qtext( std::string m_qtext );
void set_foo( std::string foo );
std::string sql();
std::string get_sql_update();
private:
std::string m_db_table; // Tabellenname
std::string m_db_key; // Tabellen-key
std::string m_qtext_where; // add.optionale where clause
std::string m_qtext; // fertiger SELECT
std::string m_sql_update; // fertiger UPDATE
bool m_erl; // nur erledigte waehlen?
std::vector<std::string> m_kids; // Liste von keys zu selecten
};
这里是一个setter方法:我用填充的字符串和向量调用它们,在这段代码中仔细检查它
void Query::setoptions( string qtext_where, bool erl, vector<string> kids ) {
m_qtext_where = qtext_where;
m_erl = erl;
m_kids = kids;
}
但是当我的应用稍后调用query.build_query()
变量是空的
Query Query::build_query( ) {
cout << "kids size" << m_kids.size() << endl;
cout << "m_qtext_where " << m_qtext_where << endl;
// Query zur auswahl der zu uebertragenden Datensaetze
string sql_update = "UPDATE " + m_db_table;
string qtext = "SELECT * FROM " + m_db_table;
string qtext_order = " ORDER BY " + m_db_key;
(...)
编辑:所以这是app代码的一部分,它调用1.setoptions和2.build_query
// read file line by line into vector of strings
vector<string> text_file;
ifstream ifs( input );
string temp;
while( getline( ifs, temp ) ) {
if (temp.substr(0,1) == "#" ) {
cout << "COMMENT: " << temp << endl;
continue;
}
cout << temp << endl;
text_file.push_back( temp );
}
// check: yes, vector has a size = number of lines
cout << "text_file size " << text_file.size() << endl;
// create Query object
Query query = Query();
// set the members, bool erl = true
query.setoptions( "", erl, text_file );
// call 2nd method
q2 = query.build_query();
答案 0 :(得分:2)
如果没有完整的代码,真的无法告诉你什么,但是我怀疑你从query.build_query返回一个查询对象,它不是查询对象的完整副本,如果这有意义的话?你能包含build_query的全文吗?
另外,我将build_query方法设为void,而不是尝试将一个新的Query对象分配回第二个Query对象(q2)(除非你真的需要,再次,如果没有完整代码),像这样:
void Query::build_query( ) {
std::cout << "kids size" << m_kids.size() << std::endl;
std::cout << "m_qtext_where " << m_qtext_where << std::endl;
}
main
{
...
Query query = Query();
// set the members, bool erl = true
query.setoptions( "", true, text_file );
// call 2nd method
query.build_query();
}
另外,在这里只是迂腐,但鉴于你为所有选项提供默认args,我倾向于在构造函数中初始化它们:
Query::Query()
: m_qtext_where("")
, qtext_erl(true)
, kids (std::vector<std::string>()
{}
然后代替setOptions方法,为每个单独的变量设置setter:
void setWhere(std::string qtext_where) {m_qtext_where = qtext_where ;}
void setErl(bool query_erl) { m_erl = query_erl; }
void setKids(std::vector<std::string> kids) { m_kids = kids; }
只有在你需要时才打电话。