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Determine if Type is a pointer in a template function
我正在寻找一种方法,确定模板是不是在汇编时间的点。 由于T不是点子,方案将无法编纂,因为你不能删除正常类型的变量。
template <typename T>
void delete(T &aVar)
{
// if T is a point
delete aVar;
aVar = 0;
// if T is not a point, do nothing
}
基本上,我正在学习自己制定联系清单(而不是使用STL清单)。 我试图在我的名单上使用模板,以便它能够采取任何形式的行动。 当类型为点人时,我要自动删除(关键词删除)。
如上所述,当我使用名单上某类人员时,VC2010年赢得了汇编,因为你不能删除任何点变量。 因此,我正在寻求一种方法,例如,在删除枪支时要先从宏观上删除,或者不按照模板类型加以汇编。
这是一种手法,但我认为,在使用本地的<代码>delete后,更宜使用NUL。
只考虑对可变点人类型论点的职能,使用
template< typename T > // The compiler may substitute any T,
void delete_ref( T *&arg ); // but argument is still a pointer in any case.
简单地发现某一类型是否为点人,使用is_pointer from <type_traits> in Boost, TR1, or C++0x。
如何将这一职能带进T*而不是T?
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