方案一
代碼如下:
class qmmanager
{
public:
static qmmanager &instance()
{
static qmmanager instance_;
return instance_;
}
}
這是最簡(jiǎn)單的版本,在單線程下(或者是c++0x下)是沒(méi)任何問(wèn)題的,但在多線程下就不行了,因?yàn)閟tatic qmmanager instance_;這句話不是線程安全的。
在局部作用域下的靜態(tài)變量在編譯時(shí),編譯器會(huì)創(chuàng)建一個(gè)附加變量標(biāo)識(shí)靜態(tài)變量是否被初始化,會(huì)被編譯器變成像下面這樣(偽代碼):
代碼如下:
static qmmanager &instance()
{
static bool constructed = false;
static uninitialized qmmanager instance_;
if (!constructed) {
constructed = true;
new(&s) qmmanager; //construct it
}
return instance_;
}
這里有競(jìng)爭(zhēng)條件,兩個(gè)線程同時(shí)調(diào)用instance()時(shí),一個(gè)線程運(yùn)行到if語(yǔ)句進(jìn)入后還沒(méi)設(shè)constructed值,此時(shí)切換到另一線程,constructed值還是false,同樣進(jìn)入到if語(yǔ)句里初始化變量,兩個(gè)線程都執(zhí)行了這個(gè)單例類的初始化,就不再是單例了。
方案二
一個(gè)解決方法是加鎖:
代碼如下:
static qmmanager &instance()
{
lock(); //鎖自己實(shí)現(xiàn)
static qmmanager instance_;
unlock();
return instance_;
}
但這樣每次調(diào)用instance()都要加鎖解鎖,代價(jià)略大。
方案三
那再改變一下,把內(nèi)部靜態(tài)實(shí)例變成類的靜態(tài)成員,在外部初始化,也就是在include了文件,main函數(shù)執(zhí)行前就初始化這個(gè)實(shí)例,就不會(huì)有線程重入問(wèn)題了:
代碼如下:
class qmmanager
{
protected:
static qmmanager instance_;
qmmanager();
~qmmanager(){};
public:
static qmmanager *instance()
{
return &instance_;
}
void do_something();
};
qmmanager qmmanager::instance_; //外部初始化
這被稱為餓漢模式,程序一加載就初始化,不管有沒(méi)有調(diào)用到。
看似沒(méi)問(wèn)題,但還是有坑,在一個(gè)2b情況下會(huì)有問(wèn)題:在這個(gè)單例類的構(gòu)造函數(shù)里調(diào)用另一個(gè)單例類的方法可能會(huì)有問(wèn)題。
看例子:
代碼如下:
//.h
class qmmanager
{
protected:
static qmmanager instance_;
qmmanager();
~qmmanager(){};
public:
static qmmanager *instance()
{
return &instance_;
}
};
class qmsqlite
{
protected:
static qmsqlite instance_;
qmsqlite();
~qmsqlite(){};
public:
static qmsqlite *instance()
{
return &instance_;
}
void do_something();
};
qmmanager qmmanager::instance_;
qmsqlite qmsqlite::instance_;
//.cpp
qmmanager::qmmanager()
{
printf(qmmanager constructorn);
qmsqlite::instance()->do_something();
}
qmsqlite::qmsqlite()
{
printf(qmsqlite constructorn);
}
void qmsqlite::do_something()
{
printf(qmsqlite do_somethingn);
}
這里qmmanager的構(gòu)造函數(shù)調(diào)用了qmsqlite的instance函數(shù),但此時(shí)qmsqlite::instance_可能還沒(méi)有初始化。
這里的執(zhí)行流程:程序開(kāi)始后,在執(zhí)行main前,執(zhí)行到qmmanager qmmanager::instance_;這句代碼,初始化qmmanager里的instance_靜態(tài)變量,調(diào)用到qmmanager的構(gòu)造函數(shù),在構(gòu)造函數(shù)里調(diào)用qmsqlite::instance(),取qmsqlite里的instance_靜態(tài)變量,但此時(shí)qmsqlite::instance_還沒(méi)初始化,問(wèn)題就出現(xiàn)了。
那這里會(huì)crash嗎,測(cè)試結(jié)果是不會(huì),這應(yīng)該跟編譯器有關(guān),靜態(tài)數(shù)據(jù)區(qū)空間應(yīng)該是先被分配了,在調(diào)用qmmanager構(gòu)造函數(shù)前,qmsqlite成員函數(shù)在內(nèi)存里已經(jīng)存在了,只是還未調(diào)到它的構(gòu)造函數(shù),所以輸出是這樣:
qmmanager constructor
qmsqlite do_something
qmsqlite constructor
方案四
那這個(gè)問(wèn)題怎么解決呢,單例對(duì)象作為靜態(tài)局部變量有線程安全問(wèn)題,作為類靜態(tài)全局變量在一開(kāi)始初始化,有以上2b問(wèn)題,那結(jié)合下上述兩種方式,可以解決這兩個(gè)問(wèn)題。boost的實(shí)現(xiàn)方式是:?jiǎn)卫龑?duì)象作為靜態(tài)局部變量,但增加一個(gè)輔助類讓單例對(duì)象可以在一開(kāi)始就初始化。如下:
代碼如下:
//.h
class qmmanager
{
protected:
struct object_creator
{
object_creator()
{
qmmanager::instance();
}
inline void do_nothing() const {}
};
static object_creator create_object_;
qmmanager();
~qmmanager(){};
public:
static qmmanager *instance()
{
static qmmanager instance;
return &instance;
}
};
qmmanager::object_creator qmmanager::create_object_;
class qmsqlite
{
protected:
qmsqlite();
~qmsqlite(){};
struct object_creator
{
object_creator()
{
qmsqlite::instance();
}
inline void do_nothing() const {}
};
static object_creator create_object_;
public:
static qmsqlite *instance()
{
static qmsqlite instance;
return &instance;
}
void do_something();
};
qmmanager::object_creator qmmanager::create_object_;
qmsqlite::object_creator qmsqlite::create_object_;
結(jié)合方案3的.cpp,這下可以看到正確的輸出和調(diào)用了:
qmmanager constructor
qmsqlite constructor
qmsqlite do_something
來(lái)看看這里的執(zhí)行流程:
初始化qmmanager類全局靜態(tài)變量create_object_
->調(diào)用object_creator的構(gòu)造函數(shù)
->調(diào)用qmmanager::instance()方法初始化單例
->執(zhí)行qmmanager的構(gòu)造函數(shù)
->調(diào)用qmsqlite::instance()
->初始化局部靜態(tài)變量qmsqlite instance
->執(zhí)行qmsqlite的構(gòu)造函數(shù),然后返回這個(gè)單例。
跟方案三的區(qū)別在于qmmanager調(diào)用qmsqlite單例時(shí),方案3是取到全局靜態(tài)變量,此時(shí)這個(gè)變量未初始化,而方案四的單例是靜態(tài)局部變量,此時(shí)調(diào)用會(huì)初始化。
跟最初方案一的區(qū)別是在main函數(shù)前就初始化了單例,不會(huì)有線程安全問(wèn)題。
最終boost
上面為了說(shuō)明清楚點(diǎn)去除了模版,實(shí)際使用是用模版,不用寫那么多重復(fù)代碼,這是boost庫(kù)的模板實(shí)現(xiàn):
代碼如下:
template <typename t>
struct singleton
{
struct object_creator
{
object_creator(){ singleton<t>::instance(); }
inline void do_nothing()const {}
};
static object_creator create_object;
public:
typedef t object_type;
static object_type& instance()
{
static object_type obj;
//據(jù)說(shuō)這個(gè)do_nothing是確保create_object構(gòu)造函數(shù)被調(diào)用
//這跟模板的編譯有關(guān)
create_object.do_nothing();
return obj;
}
};
template <typename t> typename singleton<t>::object_creator singleton<t>::create_object;
class qmmanager
{
protected:
qmmanager();
~qmmanager(){};
friend class singleton<qmmanager>;
public:
void do_something(){};
};
int main()
{
singleton<qmmanager>::instance()->do_something();
return 0;
}
其實(shí)boost庫(kù)這樣的實(shí)現(xiàn)像打了幾個(gè)補(bǔ)丁,用了一些奇技淫巧,雖然確實(shí)繞過(guò)了坑實(shí)現(xiàn)了需求,但感覺(jué)挺不好的。
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