C＆＃x2B;＆＃x2B;实例化点：要求或定义？

发布于 2025-01-29 05:05:05 字数 527 浏览 1 评论 0原文

当函数模板被隐式实例化时，是声明还是所需模板的定义？

看来编译器只需要声明：

template <typename T>
T f();

int g() {
  // Compiles OK, even though f's definition is not available here.
  return f<int>();
}

template <typename T>
T f() {
  return 123;
}

int main()
{
  std::cout << g();
}

我正在努力查看标准中指定此行为的位置。谁能帮忙找到这个？

例如，“ temp.point”条款不是说F的实例化点应立即遵循G的定义，这应该导致编译误差，因为尚未定义F？

＆lt; ...＆gt;这种专业的实例化点遵循名称空间范围声明或定义这是指专业。

原文

When a function template is implicitly instantiated, is the declaration or the definition of the template required?

It seems that compilers only require a declaration:

template <typename T>
T f();

int g() {
  // Compiles OK, even though f's definition is not available here.
  return f<int>();
}

template <typename T>
T f() {
  return 123;
}

int main()
{
  std::cout << g();
}

I'm struggling to see where this behavior is specified in the Standard. Could anyone help finding that?

For example, doesn't the "temp.point" clause say that the point of f's instantiation should immediately follow g's definition, which should lead to compilation error as the f is not defined yet?

<...> the point of instantiation for such a specialization immediately
follows the namespace scope declaration or definition
that refers to the specialization.

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寒冷纷飞旳雪 2025-02-05 05:05:05

只要有一个模板声明其专业化被推荐的程序是有效的。在您的情况下，您提供了f的模板声明，该模板可以由编译器使用来知道要拨打哪个专业化。但是，定义可以在以后提供（如您所做的），并且需要呼叫成功（而不是要求呼叫以绑定到函数）。

此外，专业化的实例化f＆lt; int将在g之后，如以下示例中使用的注释所强调的。在下面的示例中，点标记为＃1是专业f＆lt; int的POI。

还要注意，从 temp.point＃7 ：

函数模板，成员函数模板或类模板的成员函数或静态数据成员的专业化可能在翻译单元中具有多个实例化点，除了实例化点上面描述的是，对于在翻译单元中具有实例化点的任何此类专业化，翻译单元的末端也被认为是实例化的点。

这意味着将有第二个POI标记为＃2 < /code>在以下给定示例中，相同的专业化f＆lt; int＆gt;（）。

因此，＃1和＃2下面标记的都被视为专业化的POI。

template <typename T>
T f();

int g() {
  
  return f<int>(); //#0 point of call but not point of instantiation
}
// #1: This is the first POI for f<int>()

template <typename T>
T f() {
  return 123;
}

int main()
{
  std::cout << g();
}

//#2 Second POI for f<int>() at the end of TU due to temp.point#7

As long as there is a template declaration whose specialization is referred the program is valid. In your case, you have provided a template declaration for f that can be used by the compiler to know which specialization to call. The definition however can be provided later(as you did) and which is needed for the call to succeed(and not for the call to be bound to a function).

Moreover, the point of instantiation for the specialization f<int> will be just after g as highlighted in the below example using comments. In the below sample, point labeled #1 is the POI for the specialization f<int>.

Note also that from temp.point#7:

A specialization for a function template, a member function template, or of a member function or static data member of a class template may have multiple points of instantiations within a translation unit, and in addition to the points of instantiation described above, for any such specialization that has a point of instantiation within the translation unit, the end of the translation unit is also considered a point of instantiation.

This means that there will be a second POI labeled as #2 in the below given example for the same specialization f<int>().

Thus both #1 and #2 labeled below, are considered POI for the specialization.

template <typename T>
T f();

int g() {
  
  return f<int>(); //#0 point of call but not point of instantiation
}
// #1: This is the first POI for f<int>()

template <typename T>
T f() {
  return 123;
}

int main()
{
  std::cout << g();
}

//#2 Second POI for f<int>() at the end of TU due to temp.point#7

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