这是C ++的初始化列表语法的问题。
是否有可能从调用初始化列表函数,而不让其参数成员对象的构造函数?
下面列出的代码示例在工作时有类似的情况转述(paracoded?)。
情况
编码
#include <iostream>
#define LOG { std::cout << __PRETTY_FUNCTION__ << std::endl; }
namespace
{
template <class T>
class SingletonService
{
public:
static T* Instance() { LOG; return mpT; }
static void InstallInstance(T* pT) { LOG; mpT = pT; }
static void DeleteInstance() { if (mpT) delete mpT; }
protected:
static T* mpT;
};
template <class T>
T* SingletonService<T>::mpT = NULL;
class OneOfMe
{
public:
OneOfMe() { LOG; };
virtual ~OneOfMe() { };
};
class Container
{
public:
Container(OneOfMe* pObj) { LOG; /* Do something with pObj */ }
virtual ~Container() { }
};
int GenerateNum()
{
return 42;
}
class Baz
{
public:
Baz(int num) : mNum(num) { LOG; }
virtual ~Baz() { }
protected:
int mNum;
};
class Bar
{
public:
Bar() : mBaz(GenerateNum()) { LOG; } // Perfectly OK to call function that is argument to member object's non-default ctor.
virtual ~Bar() { };
protected:
Baz mBaz;
};
class Foo
{
public:
Foo()
: SingletonService<OneOfMe>::InstallInstance(new OneOfMe) // Compile error
, mContainer(SingletonService<OneOfMe>::Instance()) { }
virtual ~Foo() { };
protected:
Container mContainer;
};
}
int main(int argc, char* argv[])
{
LOG;
Bar bar;
SingletonService<OneOfMe>::InstallInstance(new OneOfMe); // This works.
Container container(SingletonService<OneOfMe>::Instance()); // And this works.
SingletonService<OneOfMe>::DeleteInstance();
return 0;
}
编译错误
>g++ main.cpp
main.cpp: In constructor ‘<unnamed>::Foo::Foo()’:
main.cpp:45: error: expected class-name before ‘(’ token
main.cpp:45: error: no matching function for call to
‘<unnamed>::Container::Container()’
main.cpp:37: note: candidates are:
<unnamed>::Container::Container(<unnamed>::OneOfMe*)
main.cpp:35: note:
<unnamed>::Container::Container(const<unnamed>::Container&)
main.cpp:45: error: expected ‘{’ before ‘(’ token
问题
从语法构造上可以从类构造函数的初始值设定项列表中调用函数而无需成为成员对象的非默认构造函数的参数吗?
现在的问题是学术的好奇心。我知道至少有其他解决方案是在创建包含类之前实例化单例。
您可以使用逗号运算符。
在你的例子中
class Foo
{
public:
Foo()
: mContainer((SingletonService<OneOfMe>::InstallInstance(new OneOfMe),
SingletonService<OneOfMe>::Instance()))
{}
virtual ~Foo();
protected:
Container mContainer;
};
请注意两个表达式之间的附加括号,否则它们将被解释为两个而不是一个参数。
解决此特定问题的另一种方法也可以是从中返回单例InstallInstance()
,例如
template <class T>
class SingletonService {
public:
static T *InstallInstance(T *pT) { LOG; return mpT = pT; }
};
然后
class Foo {
public:
Foo()
: mContainer(SingletonService<OneOfMe>::InstallInstance(new OneOfMe)) {}
virtual ~Foo();
protected:
Container mContainer;
};
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