我们可以使用lambda表达式作为函数参数的默认值吗? [英] Can we use a lambda-expression as the default value for a function argument?
问题描述
参考C ++ 11规范(5.1.2.13):
A lambda-expression 不会隐含或显式地捕获任何实体。
[示例:void f2(){
int i = 1;
void g1(int =([i] {return i;})()); //形式
void g2(int =([i] {return 0;})()); //形成的
void g3(int =([=] {return i;})()); //形成的
void g4(int =([=] {return 0;})()); // OK
void g5(int =([] {return sizeof i;})()); // OK
}
但是,我们还可以使用lambda表达式本身作为函数参数的默认值吗?
$ p
$ b
< b $ b
例如
template< typename functor>
void foo(functor const& f = [](int x){return x;})
{
}
是的。在这方面,lambda表达式与其他表达式(例如, 0
)没有区别。但请注意,扣除不使用默认参数。换句话说,如果您声明
template< typename T>
void foo(T = 0);
然后 foo(0);
foo< int>
但是 foo()
您需要显式调用 foo< int>()
。因为在你的情况下你使用一个lambda表达式,因为表达式的类型(在默认参数的站点)是唯一的,所以可以调用 foo
。但是你可以这样做:
//可能隐藏在一个细节命名空间或一些这样的
auto default_parameter = [] int x){return x; };
template<
typename Functor = decltype(default_parameter)
>
void foo(Functor f = default_parameter);
Refering to the C++11 specification (5.1.2.13):
A lambda-expression appearing in a default argument shall not implicitly or explicitly capture any entity.
[ Example:void f2() { int i = 1; void g1(int = ([i]{ return i; })()); // ill-formed void g2(int = ([i]{ return 0; })()); // ill-formed void g3(int = ([=]{ return i; })()); // ill-formed void g4(int = ([=]{ return 0; })()); // OK void g5(int = ([]{ return sizeof i; })()); // OK }
—end example ]
However, can we also use a lambda-expression itself as the default value for a function argument?
e.g.
template<typename functor>
void foo(functor const& f = [](int x){ return x; })
{
}
Yes. In this respect lambda expressions are no different from other expressions (like, say, 0
). But note that deduction is not used with defaulted parameters. In other words, if you declare
template<typename T>
void foo(T = 0);
then foo(0);
will call foo<int>
but foo()
is ill-formed. You'd need to call foo<int>()
explicitly. Since in your case you're using a lambda expression nobody can call foo
since the type of the expression (at the site of the default parameter) is unique. However you can do:
// perhaps hide in a detail namespace or some such
auto default_parameter = [](int x) { return x; };
template<
typename Functor = decltype(default_parameter)
>
void foo(Functor f = default_parameter);
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