功能模板 [英] Function templates
问题描述
1。如果定义了函数模板：
模板< typename A>
A SQ（A& a）
{
返回a * a;
}
如何阻止用户传递一个没有意义的类型，例如
员工类类型？
2.如果员工在下面的模板中作为T传递，那么是否必须定义任务
运算符因为在
函数中有赋值操作？
模板< class T>
void swap（T& x，T& y）
{
T temp;
temp = x;
x = y;
y = temp;
}
3.是" template< class T>"等同于模板< typename T>？
谢谢！
< blockquote>
" al" <人*** @ 168.net>在消息新闻中写道：ne *********************** @ bgtnsc04-news.ops.worldnet.att.net ...
1.如果定义了一个函数模板：
如何阻止用户传递一个没有意义的类型，比如
员工类类型？
你不能，你可以阻止他们写作
员工（）*员工（）
其他比编译器婊子没有为那些
类型定义。
2.如果Employee在下面的模板中作为T传递，那么应该是一个赋值
运算符必须定义，因为在
函数中有赋值操作？
必须有一个可访问。当然，如果用户没有声明一个，则将在一个类中定义operator =
。
3.是template< class T>等同于模板< typename T>？
是的。 < T类>将匹配任何类型，而不仅仅是类类型。类型名
和类在这种情况下是可以互换的。
" al" <人*** @ 168.net>在消息中写道
news：ne *********************** @ bgtnsc04-news.ops.worldnet.att.net。 ..1.如果定义了功能模板：
模板< typename A>
A SQ（A& a）
{
返回一个* a;
}
如何阻止用户传递一个没有意义的类型，比如
员工类类型？
让我们试试吧。它不会编译。当然错误信息可能有点
模糊。
2.如果员工在下面的模板中作为T传递，那么作业的运营商是否必须定义，因为在
函数中有赋值操作？
模板< class T>
void swap（T& x，T& y）
{
T temp;
temp = x;
x = y;
y = temp;
}
如果你不会编写一个赋值运算符，编译器将为你提供
只能复制这些位。
3.是template< class T> ;相当于模板< typename T>？
是的。关键字typename这是语言的一个相对较晚的补充。
在这种情况下更有意义的是类。
谢谢！
al写道：1.如果定义了功能模板：
模板< typename A>
A SQ（A& a）
{
返回a * a;
}
如何阻止用户传递一个没有意义的类型，比如
员工类类型？
你可以创建如下的专业化：
模板< typename A>
A SQ（A& a）
{
返回a * a;
}
模板<>
员工SQ（员工& a）
{
...一些编译成错误的代码...
}
但是如果你没有提供''运营商*（）''，你的原始SQ无论如何都会提供
。
1. If a function template is defined:
template <typename A>
A SQ(A& a)
{
return a*a;
}
How could prevent users from passing a type which makes no sense, like
Employee class type?
2. If Employee is pass as T in the template below, should an assignment
operator have to be defined since there''re assignment operation in the
function?
template <class T>
void swap(T& x, T& y)
{
T temp;
temp = x;
x = y;
y = temp;
}
3. Is "template <class T>" equivalent to "template <typename T>"?
Thanks!
"al" <al***@168.net> wrote in message news:ne***********************@bgtnsc04-news.ops.worldnet.att.net...1. If a function template is defined:
How could prevent users from passing a type which makes no sense, like
Employee class type?
You can''t, anymore than you can stop them from writing
Employee() * Employee()
other than the compiler bitching about * not being defined for those
types.
2. If Employee is pass as T in the template below, should an assignment
operator have to be defined since there''re assignment operation in the
function?
There has to be one accessible. Of course, operator= will be defined
in a class if the user doesn''t declare one.
3. Is "template <class T>" equivalent to "template <typename T>"?
Yes. <class T> will match any type, not just class types. Typename
and class are interchangeable in this context.
"al" <al***@168.net> wrote in message
news:ne***********************@bgtnsc04-news.ops.worldnet.att.net...1. If a function template is defined:
template <typename A>
A SQ(A& a)
{
return a*a;
}
How could prevent users from passing a type which makes no sense, like
Employee class type?
Let ''em try. It won''t compile. Of course the error message may be a bit
obscure.
2. If Employee is pass as T in the template below, should an assignment
operator have to be defined since there''re assignment operation in the
function?
template <class T>
void swap(T& x, T& y)
{
T temp;
temp = x;
x = y;
y = temp;
}
If you don''t write an assignment operator the compiler will provide you with
one which just copies the bits.
3. Is "template <class T>" equivalent to "template <typename T>"?
Yes. The keyword "typename" was a relatively late addition to the language.
It makes more sense in this context then "class".
Thanks!
al wrote:1. If a function template is defined:
template <typename A>
A SQ(A& a)
{
return a*a;
}
How could prevent users from passing a type which makes no sense, like
Employee class type?
You can create a specialization like:
template <typename A>
A SQ(A& a)
{
return a*a;
}
template <>
Employee SQ(Employee& a)
{
... some code that compiles to an error ...
}
However if you don''t provide a ''operator *()'', your original SQ provides
for an error anyway.
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