如何推导函子的返回值的类型? [英] How to deduce the type of the functor's return value?
问题描述
我想写一个接受2个值和函数或lambda的模板函数。
该函数用这些值调用函子并返回结果。
模板< typename T,typename Fn>
_ReturnTypeOfPred_ Apply(T x,T y,Fn fn)
{
return fn(x,y);问题:如何定义的返回类型?
}
$ b <应用等于 Fn
的返回类型?它不一定等于 T
,如函子
template< typename T>
auto Sum(T x,T y) - > decltype(x + y)
{
return x + y;
}
更新
第一个例子过于简单。
这个工作吗?
模板< typename TContainer,typename Fn>
auto Apply(const TContainer& x,const TContainer& y,Fn fn) - > decltype(fn(x.front(),y.front()))
{
return fn(x.front(),y.front());
}
如果我重复 code>表达式在
decltype
的返回类型?是否有更优雅的方式?
只需使用 decltype
:
template< typename T,typename Fn>
auto Apply(T x,T y,Fn fn) - > decltype(fn(x,y))
{
return fn(x,y);
}
您可以使用 std :: result_of
( std :: result_of和decltype 之间的差异),但为什么
模板< typename T,typename Fn>
typename std :: result_of< Fn,T,T> :: type Apply(T x,T y,Fn fn)
{
return fn(x,y);
}
问题:对于函数
auto fn(< args>) - > < return-type> {return< expression> ;; }
用 return-type
c $ c> decltype(< expression>)将通常工作,但可能容易出错。例如,考虑:
auto f(char c) - > decltype(std :: string()+ = c){return std :: string()+ = c; }
这里 decltype
$ c> std :: string& ,你的函数将返回一个本地的左值引用!这将必须更改为:
auto f(char c) - > std :: remove_reference< decltype(std :: string()+ = c)> :: type {
return std :: string()+ = c;
}
在其他情况下,< expression& code>可能产生一个值,由于是eg不可返回不可复制,包含lambda等。
I would like to write a template function which accepts 2 values and a functor or a lambda. The function calls the functor with those values and returns the result.
template <typename T, typename Fn>
_ReturnTypeOfPred_ Apply(T x, T y, Fn fn)
{
return fn(x, y);
}
Question: How can I define the return type of Apply
to become equal to the return type of Fn
? It is not necessarily equal to T
, as in this example of the functor
template <typename T>
auto Sum(T x, T y) -> decltype(x+y)
{
return x+y;
}
Update
The 1st example was oversimplified. Should this one work?
template <typename TContainer, typename Fn>
auto Apply(const TContainer& x, const TContainer& y, Fn fn) -> decltype(fn(x.front(), y.front()))
{
return fn(x.front(), y.front());
}
Would it always work if I repeat return
expression in decltype
of the return type? Is there a more elegant way?
You're nearly there; just use decltype
:
template <typename T, typename Fn>
auto Apply(T x, T y, Fn fn) -> decltype(fn(x, y))
{
return fn(x, y);
}
You could use std::result_of
(Difference between std::result_of and decltype) but why bother?
template <typename T, typename Fn>
typename std::result_of<Fn, T, T>::type Apply(T x, T y, Fn fn)
{
return fn(x, y);
}
Regarding the follow-up question: for a function
auto fn(<args>) -> <return-type> { return <expression>; }
substituting return-type
with decltype(<expression>)
will usually work, but can be error prone. For example, consider:
auto f(char c) -> decltype(std::string() += c) { return std::string() += c; }
Here decltype
will yield std::string &
and your function will return an lvalue reference to a local! This would have to be changed to:
auto f(char c) -> std::remove_reference<decltype(std::string() += c)>::type {
return std::string() += c;
}
In other cases, <expression>
could yield a value that is not returnable for reason of being e.g. noncopyable, containing a lambda, etc.
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