什么时候是“typename"需要关键字吗?
可能的重复:
正式来说,typename 是做什么用的?
我必须将模板放在哪里以及为什么和 typename 关键字?
考虑下面的代码:
template<class K>
class C {
struct P {};
vector<P> vec;
void f();
};
template<class K> void C<K>::f() {
typename vector<P>::iterator p = vec.begin();
}
为什么在这个例子中需要typename"关键字?还有其他必须指定typename"的情况吗?
Why is the "typename" keyword necessary in this example? Are there any other cases where "typename" must be specified?
推荐答案
简短回答:每当引用作为依赖名称的嵌套名称时,即嵌套在具有未知参数的模板实例中.
Short answer: Whenever referring to a nested name that is a dependent name, i.e. nested inside a template instance with unknown parameter.
长答案:C++ 中有三层实体:值、类型和模板.所有这些都可以有名称,仅凭名称并不能告诉您它是实体的哪一层.相反,有关名称实体性质的信息必须从上下文中推断出来.
Long answer: There are three tiers of entities in C++: values, types, and templates. All of those can have names, and the name alone doesn't tell you which tier of entity it is. Rather, the information about the nature of a name's entity must be inferred from the context.
当此推断不可能时,您必须指定它:
Whenever this inference is impossible, you have to specify it:
template <typename> struct Magic; // defined somewhere else
template <typename T> struct A
{
static const int value = Magic<T>::gnarl; // assumed "value"
typedef typename Magic<T>::brugh my_type; // decreed "type"
// ^^^^^^^^
void foo() {
Magic<T>::template kwpq<T>(1, 'a', .5); // decreed "template"
// ^^^^^^^^
}
};
这里的名称 Magic
、Magic
和 Magic
必须加以解释,因为无法判断:由于 Magic
是一个模板,Magic
类型的 性质 取决于在 T
上――例如,可能存在与主模板完全不同的特化.
Here the names Magic<T>::gnarl
, Magic<T>::brugh
and Magic<T>::kwpq
had to be expliciated, because it is impossible to tell: Since Magic
is a template, the very nature of the type Magic<T>
depends on T
-- there may be specializations which are entirely different from the primary template, for example.
使 Magic
成为依赖名称的原因是我们位于模板定义中,其中 T
是未知的.如果我们使用 Magic
,情况会有所不同,因为编译器知道(你保证!)Magic
的完整定义.
What makes Magic<T>::gnarl
a dependent name is the fact that we're inside a template definition, where T
is unknown. Had we used Magic<int>
, this would be different, since the compiler knows (you promise!) the full definition of Magic<int>
.
(如果您想自己测试,这里有一个您可以使用的 Magic
示例定义.请原谅在专业化中为简洁起见使用 constexpr
;如果您有一个旧的编译器,可以随意将静态成员常量声明更改为旧式 pre-C++11 形式.)
(If you want to test this yourself, here's a sample definition of Magic
that you can use. Pardon the use of constexpr
in the specializaation for brevity; if you have an old compiler, feel free to change the static member constant declaration to the old-style pre-C++11 form.)
template <typename T> struct Magic
{
static const T gnarl;
typedef T & brugh;
template <typename S> static void kwpq(int, char, double) { T x; }
};
template <> struct Magic<signed char>
{
// note that `gnarl` is absent
static constexpr long double brugh = 0.25; // `brugh` is now a value
template <typename S> static int kwpq(int a, int b) { return a + b; }
};
用法:
int main()
{
A<int> a;
a.foo();
return Magic<signed char>::kwpq<float>(2, 3); // no disambiguation here!
}
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