我正在阅读本文page中的这篇文章,所以让我提出代码:
template <class T> class D: typename C<T>::Base { //#1 error
D(): typename C<T>::Base(typename C<T>::Base(0)) { } //#2 error, #3 correct
typename C<T> f() { //#4 error
typename C<T>::A* p; //#5 correct
A<T>::B * n;
}
};
class Q{
typename C<long>::A * p; // #6 error
}
template <class T, class U> class R{
typename C<long>::A * p; // #7 optional
}
#3
是正确的,但我试图了解作者试图传达的内容。他说:
typename#3:正确。这里,typename用于消除歧义 参数类型。没有typename,表达式C :: Base(0) 将被视为对名为Base的函数的调用。随着 typename前缀,C :: Base(0)创建一个类型的临时对象 C :: Base使用参数0初始化。
另外,如果您在该部分上方看到一点,作者说:
typename关键字必须在该名称前面加上一个从属名称 满足以下三条规则:
1.它出现在模板中
2. 这是合格的//我完全无法理解这一行与此引语的起始段
3.它不用于基类规范或成员初始化 列表。
我根本无法理解这一行(上面的#2)以及此引文的起始段落。你能解释一下作者的意思吗?
答案 0 :(得分:2)
“合格”意味着它不在同一范围内,而是当前范围的子范围,也取决于模板。例如,
template <class T> struct C {
typedef int b;
}
b
在另一个模板中引用时是合格的id
,例如
template<typename T> struct M {
typename C<T>::b bb;
}
标准的相关引用(§14.6.3):
引用类型的限定ID,其中包含 nested-name-specifier依赖于template-parameter(14.6.2) 以关键字typename为前缀以指示qualified-id 表示一种类型,形成一个详细类型说明符(7.1.5.3)。
和§14.6.2:
A name used in a template declaration or definition and that is dependent on a template-parameter is
assumed not to name a type unless the applicable name lookup finds a type name or the name is qualified
by the keyword typename. [Example:
// no B declared here
class X;
template<class T> class Y {
class Z; // forward declaration of member class
void f() {
X* a1; // declare pointer to X
T* a2; // declare pointer to T
Y* a3; // declare pointer to Y<T>
Z* a4; // declare pointer to Z
typedef typename T::A TA;
TA* a5; // declare pointer to T’s A
typename T::A* a6; // declare pointer to T’s A
T::A* a7; // T::A is not a type name:
// multiply T::A by a7; ill-formed,
// no visible declaration of a7
B* a8; // B is not a type name:
// multiply B by a8; ill-formed,
// no visible declarations of B and a8
}
};
—end example]
和§14.6.7:
Within the definition of a class template or within the definition of a member of a class template, the keyword
typename is not required when referring to the unqualified name of a previously declared member
of the class template that declares a type. The keyword typename shall always be specified when the
member is referred to using a qualified name, even if the qualifier is simply the class template name.
[Example:
template<class T> struct A {
typedef int B;
A::B b; // ill-formed: typename required before A::B
void f(A<T>::B); // ill-formed: typename required before A<T>::B
typename A::B g(); // OK
};
The keyword typename is required whether the qualified name is A or A<T> because A or A<T> are synonyms
within a class template with the parameter list <T>. ]
答案 1 :(得分:0)
不确定这是否有帮助,但是模板使用的示例:
template<typename TemplateVar>
class SimpleTemplate
{
public:
TemplateVar Item;
void SetItem(TemplateVar &ItemCopy);
TemplateVar &GetItem();
};
template<typename TemplateVar>
void SimpleTemplate<TemplateVar>::SetItem(TemplateVar &ItemCopy)
{
Item = ItemCopy;
return;
}
template<typename TemplateVar>
TemplateVar &SimpleTemplate<TemplateVar>::GetItem()
{
return Item;
}
int main(int ArgC, char *ArgV[])
{
SimpleTemplate<char> Test;
return 0;
}
随意使用代码做任何你想做的事情(如果可以的话,甚至卖掉它)。