C ++单例与私有构造函数

本文介绍了C ++单例与私有构造函数的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我需要单一的应用程序生命周期，保证创建/销毁和静态访问它。

#include< iostream> ; #include< cstdlib> #define DISALLOW_COPY_AND_ASSIGN（TypeName）\ TypeName（const TypeName&）; \ void operator =（const TypeName&） #define M（）C :: sM（） #define M2（）C :: sM2 $ b using namespace std; class C { private： static C * s; 〜C（）{cout< 〜C（）< endl; } static C * instance（）{ if（s == NULL）{s = new C（）; } cout<< instance（）=< s<< endl;返回s; } static void cleanUp（）{delete s; } void m（）{cout<< m（）< endl; } void m2（）{cout<< m2（）< endl; } DISALLOW_COPY_AND_ASSIGN（C）; public： C（）{ cout<< C（）< endl; if（s == NULL）{ s = this; atexit（& cleanUp）; cout<< cleanUp is installed< endl; } else { cout<< cleanUp未安装< endl; } } void * operator new（size_t sz）{ void * p = NULL; if（s == NULL）{p = new char [sz]; } else {p = s; } cout<< new（<< sz<<）=< p < endl; return p; } void operator delete（void * p，size_t sz）{ cout< delete（<< sz<<，<< p<<）< endl; if（p）delete [] static_cast< char *>（p）; s = NULL; } void static sM（）{cout<< sM（）< endl; instance（） - > m（）; } void static sM2（）{cout< sM2（）< endl; instance（） - > m2（）; } }; C * C :: s = NULL; int main（）{ M（）; M2（）; C * p1 = new C（）; C * p2 = new C（）; }

但我不知道如何摆脱g ++警告： p>

test.cpp：在静态成员函数'static void C :: operator delete（void *，size_t）'： test.cpp：22：warning：删除'void *'未定义

of void *，析构函数开始在无限循环中调用自身。任何人都可以帮助我获得干净的代码，没有警告？

要删除的类型是 char * c>：

void operator delete（void * p，size_t sz）{ cout< delete（<< sz<<，<< p<<）< endl; if（p）delete（char *）p; }

您的新版本中有一个错误：新的char [10]和新的char（10）之间的区别

也许应该是：

p = new char [sz];

，然后

code> delete []（char *）p;

----编辑：

纯粹主义的删除，牺牲人们学习的清晰度; P：

delete [] static_cast< char *>（p）;

（*我真的很感激关于演员表的注意事项）

I need singleton with a application lifetime, guaranteed creation/destruction and static access to it.

#include <iostream> #include <cstdlib> #define DISALLOW_COPY_AND_ASSIGN(TypeName) \ TypeName(const TypeName&); \ void operator=(const TypeName&) #define M() C::sM() #define M2() C::sM2() using namespace std; class C { private: static C* s; ~C() { cout << "~C()" << endl; } static C* instance() { if (s==NULL) { s=new C(); } cout << "instance()=" << s << endl; return s; } static void cleanUp() { delete s; } void m() { cout << "m()" << endl; } void m2() { cout << "m2()" << endl; } DISALLOW_COPY_AND_ASSIGN(C); public: C() { cout << "C()" << endl; if (s==NULL) { s=this; atexit(&cleanUp); cout << "cleanUp is installed" << endl; } else { cout << "cleanUp is not installed" << endl; } } void* operator new(size_t sz) { void* p=NULL; if (s==NULL) { p=new char[sz]; } else { p=s; } cout << "new(" << sz << ")=" << p << endl; return p; } void operator delete(void* p, size_t sz) { cout << "delete(" << sz << "," << p << ")" << endl; if (p) delete[] static_cast<char*>(p); s=NULL; } void static sM() { cout << "sM()" << endl; instance()->m(); } void static sM2() { cout << "sM2()" << endl; instance()->m2(); } }; C* C::s = NULL; int main() { M(); M2(); C* p1 = new C(); C* p2 = new C(); }

But I don't know how to get rid of g++ warning:

test.cpp: In static member function 'static void C::operator delete(void*, size_t)': test.cpp:22: warning: deleting 'void*' is undefined

If I write C* instead of void*, destructor start calling itself in infinite loop. Can anybody help me to get clean code without warnings? C++98 of course.

解决方案

The type to delete is char*:

void operator delete(void* p, size_t sz) { cout << "delete(" << sz << "," << p << ")" << endl; if (p) delete (char*) p; }

There is a bug in your new: What's the difference between new char[10] and new char(10)

Maybe should be:

p=new char[sz];

and then

delete[] (char*)p ;

---- edited:

The delete for purists, sacrificing clarity to people learning ;P :

delete[] static_cast<char*>(p) ;

(*actually I appreciate the note about the cast)